How do you solve $2x^2 - 3x = 7$ using the quadratic formula?

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Answer 1 · 2016-03-22T15:54:26

$x = -1/2+sqrt(5)$ or $x = -1/2-sqrt(5)$

Begin by setting the equation equal to zero.

$2x^2-3x -7=0$

Identify the a, b, and c terms

$ax^2-bx -c=0$
$a = 2$
$b = -3$
$c = -7$

$x = (-b(+/-)sqrt(b^2-4ac))/(2a)$

$x = (-2(+/-)sqrt(2^2-4(-3)(-7))/(2(2)))$

$x = (-2(+/-)sqrt(4-84))/(4)$

$x = (-2(+/-)sqrt(-80))/(4)$ simplify the radical $i = sqrt(-1)$

$x = (-2(+/-)4isqrt(5))/(4)$ simplify the fractions

$x = -1/2+sqrt(5)$ or $x = -1/2-sqrt(5)$

How do you solve 2x^2 - 3x = 72x^2 - 3x = 7 using the quadratic formula?