How do you solve $5 x^{2} - 2 x - 6 = - 3 x^{2} + 6 x$ $5x^2-2x-6=-3x^2+6x$ ?

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Answer 1 · 2017-06-27T10:17:01

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

First, let's add $3 x^{2}$ $3x^2$ to both sides

$8 x^{2} - 2 x - 6 = 6 x$ $8x^2-2x-6=6x$

Then, subtract $6 x$ $6x$ from both sides

$8 x^{2} - 8 x - 6 = 0$ $8x^2-8x-6=0$

Divide the entire equation by $2$ $2$

$4 x^{2} - 4 x - 3 = 0$ $4x^2-4x-3=0$

Factorise

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

Set each factor equal to $0$ $0$

$2 x + 1 = 0$ $2x+1=0$
$2 x = - 1$ $2x=-1$

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

$2 x - 3 = 0$ $2x-3=0$
$2 x = 3$ $2x=3$
$x = \frac{1}{2}$ $x=1/2$

Here is the graph

graph{4x^2-4x-3 [-3.08, 3.08, -1.54, 1.54]}

How do you solve 5x2−2x−6=−3x2+6x5x^2-2x-6=-3x^2+6x?