How do you solve $x^{2} = 8 - 5 x$ $x^2= 8- 5x$ ?

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How do you solve $x^{2} = 8 - 5 x$ $x^2= 8- 5x$ ?

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Answer 1 · 2017-06-14T05:03:18

$x = \frac{- 5 \pm \sqrt{57}}{2}$ $x = (-5 +- sqrt(57))/2$

Explanation:

The first thing to do is to convert the equation to standard form $y = a x^{2} + b x + c$ $y = ax^2 + bx + c$ . In this case of our equation, it is $x^{2} + 5 x - 8 = 0$ $x^2 + 5x - 8 = 0$ .

By using this, we know the following
$a = 1$ $a = 1$
$b = 5$ $b = 5$
$c = - 8$ $c = -8$

In order to find the value of $x$ $x$ in the equation above, we need to substitute these values to the quadratic formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/ (2a)$

In this case

$x = \frac{- 5 \pm \sqrt{5^{2} - 4 (1) (- 8)}}{2 (1)}$ $x = (-5 +- sqrt(5^2 - 4(1)(-8)))/ (2(1))$
$x = \frac{- 5 \pm \sqrt{25 - (- 32)}}{2}$ $x = (-5 +- sqrt(25 - (-32)))/2$
$x = \frac{- 5 \pm \sqrt{57}}{2}$ $x = (-5 +- sqrt(57))/2$

So we know that $x = \frac{- 5 \pm \sqrt{57}}{2}$ $x = (-5 +- sqrt(57))/2$

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How do you solve $x^{2} = 8 - 5 x$ $x^2= 8- 5x$ ?

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How do you solve $x^{2} = 8 - 5 x$ $x^2= 8- 5x$ ?