How do you solve $2 x^{2} + x - 3 = 0$ $2x^2+x-3=0$ ?

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Answer 1 · 2016-11-27T00:25:40

$x = 1, x = - \frac{3}{2}$ $x=1,x=-3/2$

$2 x^{2} + x - 3 = 0$ $2x^2+x-3=0$

Solve by factoring:
$2 x^{2} - 2 x + 3 x - 3 = 0$ $2x^2-2x+3x-3=0$

$2 x (x - 1) + 3 (x - 1) = 0$ $2x(x-1)+3(x-1)=0$

$(2 x + 3) (x - 1) = 0$ $(2x+3)(x-1)=0$

$2 x + 3 = 0$ $2x+3=0$
$2 x = - 3$ $2x=-3$
$x = - \frac{3}{2}$ $x=-3/2$

$x - 1 = 0$ $x-1=0$
$x = 1$ $x=1$

How do you solve 2x^2+x-3=02x2+x−3=02x^2+x-3=0?