How do you solve $3 x^{2} + 10 x = 8$ $3x^2 + 10x = 8$ by completing the square?

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Answer 1 · 2016-05-12T21:22:48

$x = \frac{2}{3}$ $x=2/3$ or $x = - 4$ $x=-4$

Multiply by $3$ $3$ and add $25$ $25$ to both sides to get:

$9 x^{2} + 30 x + 25 = 24 + 25$ $9x^2+30x+25=24+25$

which simplifies to:

${(3 x + 5)}^{2} = 49 = 7^{2}$ $(3x+5)^2 = 49 = 7^2$

So $(3 x + 5)$ $(3x+5)$ is a square root of $7^{2}$ $7^2$ , so it must be $7$ $7$ or $- 7$ $-7$ :

$3 x + 5 = \pm 7$ $3x+5 = +-7$

If $3 x + 5 = 7$ $3x+5 = 7$ , then $3 x = 2$ $3x=2$ , so $x = \frac{2}{3}$ $x=2/3$

If $3 x + 5 = - 7$ $3x+5 = -7$ , then $3 x = - 12$ $3x=-12$ , so $x = - 4$ $x=-4$

How do you solve 3x2+10x=83x^2 + 10x = 8 by completing the square?