How do you solve $\frac{1}{5} x^{2} - \frac{29}{25} x - \frac{6}{25} = 0$ $\frac { 1} { 5} x ^ { 2} - \frac { 29} { 25} x - \frac { 6} { 25} = 0$ ?

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Answer 1 · 2017-07-20T07:04:07

$x = 6$ $x=6$ or $x = - \frac{1}{5}$ $x=-1/5$

Step 1) First, let's make all the fractions have the same denominators:

$\frac{1}{5} x^{2} - \frac{29}{25} x - \frac{6}{25} = 0$ $1/5x^2-29/25x-6/25=0$

Becomes,

$\frac{5}{25} x^{2} - \frac{29}{25} x - \frac{6}{25} = 0$ $5/25x^2-29/25x-6/25=0$

Step 2) Then, make them as one fraction

$\frac{5 x^{2} - 29 x - 6}{25} = 0$ $(5x^2-29x-6)/25=0$

Step 3) Multiply both sides of the equation by $25$ $25$

$5 x^{2} - 29 x - 6 = 0$ $5x^2-29x-6=0$

Step 4) Factor

$5 x^{2} - 29 x - 6 = 0$ $5x^2-29x-6=0$

Becomes,

$(x - 6) (5 x + 1) = 0$ $(x-6)(5x+1)=0$

Step 5) Solve for x
$x = 6$ $x=6$ or $x = - \frac{1}{5}$ $x=-1/5$

How do you solve \frac { 1} { 5} x ^ { 2} - \frac { 29} { 25} x - \frac { 6} { 25} = 015x2−2925x−625=0\frac { 1} { 5} x ^ { 2} - \frac { 29} { 25} x - \frac { 6} { 25} = 0?