Question

$x = - b + - \sqrt{b {}^{2} } - 4ac \div 2a$
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Answer 1

The roots of the equation are $\frac{14}{15}$ and $-\frac{4}{3}$.

Step-by-step explanation:

The expression is:

$15x^{2}-28=x$

The following equation is a quadratic equation.

$15x^{2}-x-28=0$

The roots of a quadratic equation is given by:

$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a }$

Here,

a = 15

b = -1

c = -28

Compute the roots of the equation as follows:

$x = \frac{ -(-1) \pm \sqrt{(-1)^2 - 4(15)(-28)}}{ 2(15) }\\x = \frac{ 1 \pm \sqrt{1 -( -1680)}}{ 30 }\\x = \frac{ 1 \pm \sqrt{1681}}{ 30 }\\x = \frac{ 1 \pm 41\, }{ 30 }\\x = \frac{ 42 }{ 30 } \; \; ,\; -\frac{ 40 }{ 30 }\\x = \frac{ 14 }{ 15 } \; \; ,\; -\frac{ 4 }{ 3 }$

Thus, the roots of the equation are $\frac{14}{15}$ and $-\frac{4}{3}$.