How do you solve $x^{2} + 10 x + 9 = 0$ $x^2+10x+9=0$ using the quadratic formula?

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Answer 1 · 2014-10-25T17:57:01

$x = \frac{- b \pm \sqrt[2]{b^{2} - 4 a c}}{2 a}$ $x = (-b +- root 2 (b^2 - 4ac))/(2a)$

where $a$ $a$ , $b$ $b$ , and $c$ $c$ are the coefficients of the terms of the quadratic equation

For your equation we have

$x^{2} + 10 x + 9 = 0$ $x^2 + 10x + 9 = 0$

$\Rightarrow x = \frac{- 10 \pm \sqrt[2]{10^{2} - 4 (1) (9)}}{2 (1)}$ $=> x = (-10 +- root 2 (10^2 - 4(1)(9)))/(2(1))$

$\Rightarrow x = \frac{- 10 \pm \sqrt[2]{100 - 36}}{2}$ $=> x = (-10 +- root 2 (100 - 36))/(2)$

$\Rightarrow x = \frac{- 10 \pm \sqrt[2]{64}}{2}$ $=> x = (-10 +- root 2 64) / 2$

$\Rightarrow x = \frac{- 10 \pm 8}{2}$ $=> x = (-10 +- 8)/2$

$\Rightarrow x = \frac{- 10 + 8}{2}, x = \frac{- 10 - 8}{2}$ $=> x = (-10 + 8)/2, x = (-10 - 8)/2$

$\Rightarrow x = - 1, x = - 9$ $=> x = -1, x = -9$

How do you solve x^2+10x+9=0x2+10x+9=0x^2+10x+9=0 using the quadratic formula?