How do you solve $- 2 x^{2} + 7 = 2 x$ $-2x ^ { 2} + 7= 2x$ ?

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Answer 1 · 2017-05-02T07:42:40

$x = \frac{- 1}{2} \pm \frac{\sqrt{15}}{2}$ $x=(-1)/(2)±(sqrt(15))/(2)$

$- 2 x^{2} + 7 = 2 x$ $-2x^2+7=2x$

subtract $2 x$ $2x$ from both sides and set the equation equal to $0$ $0$ :
$- 2 x^{2} - 2 x + 7 = 0$ $-2x^2-2x+7=0$

plug the values $(a = - 2, b = - 2, c = 7)$ $(a=-2, b=-2, c=7)$ into the quadratic formula:

$x = \frac{- (- 2) \pm \sqrt{- 2^{2} - 4 (- 2) (7)}}{2 (- 2)}$ $x=(-(-2)±sqrt(-2^2-4(-2)(7)))/(2(-2))$

$x = \frac{2 \pm \sqrt{60}}{- 4}$ $x=(2±sqrt(60))/(-4)$

The discriminant $b 2 - 4 a c > 0$ $b2−4ac>0$
so, there are two real roots.

$x = \frac{2 \pm 2 \sqrt{15}}{- 4}$ $x=(2±2sqrt(15))/(−4)$

$x = \frac{2}{- 4} \pm \frac{2 \sqrt{15}}{- 4}$ $x=(2)/(-4)±(2sqrt(15))/(−4)$

$\to x = \frac{- 1}{2} \pm \frac{\sqrt{15}}{2}$ $→x=(-1)/(2)±(sqrt(15))/(2)$

How do you solve −2x2+7=2x-2x ^ { 2} + 7= 2x?