Show Quadratic Formula CalculatorWhat do you want to calculate?Example: 2x^2-5x-3=0 Step-By-Step ExampleLearn step-by-step how to use the quadratic formula! Example (Click to try)2x2−5x−3=0 About the quadratic formulaSolve an equation of the form ax2+bx+c=0 by using the quadratic formula: x=
Quadratic Formula Video Lesson Solve with the Quadratic Formula Step-by-Step [1:29] Need more problem types? Try MathPapa Algebra Calculator Calculator UseThis online calculator is a quadratic equation solver that will solve a second-order polynomial equation such as ax2 + bx + c = 0 for x, where a ≠ 0, using the quadratic formula. The calculator solution will show work using the quadratic formula to solve the entered equation for real and complex roots. Calculator determines whether the discriminant \( (b^2 - 4ac) \) is less than, greater than or equal to 0. When \( b^2 - 4ac = 0 \) there is one real root. When \( b^2 - 4ac > 0 \) there are two real roots. When \( b^2 - 4ac < 0 \) there are two complex roots. Quadratic Formula:The quadratic formula \( x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \) is used to solve quadratic equations where a ≠ 0 (polynomials with an order of 2) \( ax^2 + bx + c = 0 \) Examples using the quadratic formulaExample 1: Find the Solution for \( x^2 + -8x + 5 = 0 \), where a = 1, b = -8 and c = 5, using the Quadratic Formula. \( x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \) \( x = \dfrac{ -(-8) \pm \sqrt{(-8)^2 - 4(1)(5)}}{ 2(1) } \) \( x = \dfrac{ 8 \pm \sqrt{64 - 20}}{ 2 } \) \( x = \dfrac{ 8 \pm \sqrt{44}}{ 2 } \) The discriminant \( b^2 - 4ac > 0 \) so, there are two real roots. Simplify the Radical: \( x = \dfrac{ 8 \pm 2\sqrt{11}\, }{ 2 } \) \( x = \dfrac{ 8 }{ 2 } \pm \dfrac{2\sqrt{11}\, }{ 2 } \) Simplify fractions and/or signs: \( x = 4 \pm \sqrt{11}\, \) which becomes \( x = 7.31662 \) \( x = 0.683375 \) Example 2: Find the Solution for \( 5x^2 + 20x + 32 = 0 \), where a = 5, b = 20 and c = 32, using the Quadratic Formula. \( x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \) \( x = \dfrac{ -20 \pm \sqrt{20^2 - 4(5)(32)}}{ 2(5) } \) \( x = \dfrac{ -20 \pm \sqrt{400 - 640}}{ 10 } \) \( x = \dfrac{ -20 \pm \sqrt{-240}}{ 10 } \) The discriminant \( b^2 - 4ac < 0 \) so, there are two complex roots. Simplify the Radical: \( x = \dfrac{ -20 \pm 4\sqrt{15}\, i}{ 10 } \) \( x = \dfrac{ -20 }{ 10 } \pm \dfrac{4\sqrt{15}\, i}{ 10 } \) Simplify fractions and/or signs: \( x = -2 \pm \dfrac{ 2\sqrt{15}\, i}{ 5 } \) which becomes \( x = -2 + 1.54919 \, i \) \( x = -2 - 1.54919 \, i \) calculator updated to include full solution for real and complex roots EMBEDMake your selections below, then copy and paste the code below into your HTML source. ThemeOutput TypeLightbox Widget controls
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Hope that helps! You're welcome! Let me take a look... You'll be able to enter math problems once our session is over. Step-by-Step Examples Algebra Functions Find the Roots (Zeros) Step 1 Set equal to . Step 2 Solve for . Tap for more steps... Set the equal to . Solve for . Tap for more steps... Add to both sides of the equation. Divide each term in by and simplify. Tap for more steps... Divide each term in by . Simplify the left side. Tap for more steps... Cancel the common factor of . Tap for more steps... Cancel the common factor. Divide by . Step 3 Enter YOUR Problem
Mathway requires javascript and a modern browser. How do you find the zeros of a quadratic function?The zeros of a parabola are the points on the parabola that intersect the line y = 0 (the horizontal x-axis). Since these points occur where y = 0, the zeros of a quadratic function occur where f(x) = 0, or at the x-values that make ax2+bx+c=0 a x 2 + b x + c = 0 a true equation.
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