How do you solve $15 x^{2} = 7 x + 2$ $15x^2=7x+2$ ?

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How do you solve $15 x^{2} = 7 x + 2$ $15x^2=7x+2$ ?

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Answer 1 · 2018-05-28T18:13:10

$x_{1} = - \frac{1}{5}$ $x_1=-1/5$ and $x_{2} = \frac{2}{3}$ $x_2=2/3$

Explanation:

$15 x^{2} = 7 x + 2$ $15x^2=7x+2$

$15 x^{2} - 7 x - 2 = 0$ $15x^2-7x-2=0$

$15 x^{2} - 10 x + 3 x - 2 = 0$ $15x^2-10x+3x-2=0$

$5 x \cdot (3 x - 2) + 3 x - 2 = 0$ $5x*(3x-2)+3x-2=0$

$(3 x - 2) \cdot (5 x + 1) = 0$ $(3x-2)*(5x+1)=0$

So $x_{1} = - \frac{1}{5}$ $x_1=-1/5$ and $x_{2} = \frac{2}{3}$ $x_2=2/3$

Answer 2 · 2018-05-28T18:24:20

$x_{1} = \frac{2}{3}$ $x_1=2/3$ or $x_{2} = - \frac{1}{5}$ $x_2=-1/5$

Explanation:

Dividing by $15$ $15$ we get
$x^{2} - \frac{7}{15} x - \frac{2}{15} = 0$ $x^2-7/15x-2/15=0$
$x_{1, 2} = \frac{7}{20} \pm \sqrt{\frac{49}{300} + \frac{2}{15}}$ $x_{1,2}=7/20pmsqrt(49/300+2/15)$
$x_{1, 2} = \frac{7}{30} \pm \frac{13}{30}$ $x_{1,2}=7/30pm13/30$
$x_{1} = \frac{2}{3}$ $x_1=2/3$
$x_{2} = - \frac{1}{5}$ $x_2=-1/5$

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How do you solve $15 x^{2} = 7 x + 2$ $15x^2=7x+2$ ?

2 Answers

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