How do you solve $x (2 x + 3) = 9$ $x(2x + 3) = 9$ ?

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How do you solve $x (2 x + 3) = 9$ $x(2x + 3) = 9$ ?

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Answer 1 · 2016-04-15T03:49:23

-3 and $\frac{3}{2}$ $3/2$

Explanation:

Bring the equation to standard form:
$y = 2 x^{2} + 3 x - 9 = 0$ $y = 2x^2 + 3x - 9 = 0$
Use the improved quadratic formula to solve this equation.
$D = d^{2} = b^{2} - 4 a c = 9 + 72 = 81$ $D = d^2 = b^2 - 4ac = 9 + 72 = 81$ ---> $d = \pm 9$ $d = +- 9$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{3}{4} \pm \frac{9}{4} = \frac{- 3 \pm 9}{4}$ $x = -b/(2a) +- d/(2a) = -3/4 +- 9/4 = (-3 +- 9)/4$
$x = - \frac{12}{4} = - 3$ $x = -12/4 = - 3$
$x = \frac{6}{4} = \frac{3}{2}$ $x = 6/4 = 3/2$

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How do you solve $x (2 x + 3) = 9$ $x(2x + 3) = 9$ ?

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How do you solve $x (2 x + 3) = 9$ $x(2x + 3) = 9$ ?