How do you solve $2 x^{2} - 5 x = - 7$ $2x^2-5x=-7$ using the quadratic formula?

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How do you solve $2 x^{2} - 5 x = - 7$ $2x^2-5x=-7$ using the quadratic formula?

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Answer 1 · 2016-02-16T17:07:12

Solution is $\frac{5}{4} \pm i \frac{\sqrt{31}}{4}$ $5/4+-isqrt(31)/4$

Explanation:

Quadratic formula $\frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $(-b+-sqrt(b^2-4ac))/(2a)$ gives us the solution of equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ . As the given equation is $2 x^{2} - 5 x = - 7$ $2x^2−5x=−7$ or $2 x^{2} - 5 x + 7 = 0$ $2x^2−5x+7=0$ i.e. $a = 2$ $a=2$ , $b = - 5$ $b=-5$ and $c = 7$ $c=7$ , hence solution is given by

$\frac{- (- 5) \pm \sqrt{{(- 5)}^{2} - 4 \cdot 2 \cdot 7}}{2 \cdot 2}$ $(-(-5)+-sqrt((-5)^2-4*2*7))/(2*2)$ i.e.

$\frac{5 \pm \sqrt{25 - 56}}{4}$ $(5+-sqrt(25-56))/4$

$\frac{5}{4} \pm i \frac{\sqrt{31}}{4}$ $5/4+-isqrt(31)/4$

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How do you solve $2 x^{2} - 5 x = - 7$ $2x^2-5x=-7$ using the quadratic formula?

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