How do you solve $x^{2} - 5 x = - 9$ $x^2-5x= -9$ using the quadratic formula?

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Answer 1 · 2016-10-07T20:09:02

2 complex roots, $x = \frac{5 \pm 3 i}{2}$ $x=(5+-3i)/(2)$

Original Equation

$x^{2} - 5 x = - 9 \Rightarrow x^{2} - 5 x + 9 = 0$ $x^2-5x=-9 => x^2-5x+9=0$

Standard Form of a quadratic equation

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$a = 1$ $a=1$
$b = - 5$ $b=-5$
$c = 9$ $c=9$

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

$x = \frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 (1) (9)}}{2 (1)}$ $x=(-(-5)+-sqrt((-5)^2-4(1)(9)))/(2(1))$

$x = \frac{5 \pm \sqrt{25 - 36}}{2}$ $x=(5+-sqrt(25-36))/(2)$

$x = \frac{5 \pm \sqrt{- 9}}{2}$ $x=(5+-sqrt(-9))/(2)$

$x = \frac{5 \pm 3 i}{2}$ $x=(5+-3i)/(2)$

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How do you solve x2−5x=−9x^2-5x= -9 using the quadratic formula?