How do you solve $7 y^{2} - 5 y - 8 = 0$ $7y^2-5y-8=0$ using the quadratic formula?

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Answer 1 · 2016-03-16T10:00:32

$y_{1} = \frac{+ 5 + \sqrt{249}}{14}$ $y_1=(+5+sqrt(249))/(14)$ and $y_{2} = \frac{+ 5 - \sqrt{249}}{14}$ $y_2=(+5-sqrt(249))/(14)$

$y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $y=(-b+-sqrt(b^2-4ac))/(2a)$

Let $a = 7$ $a=7$ and $b = - 5$ $b=-5$ and $c = - 8$ $c=-8$

$y = \frac{- - 5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot 7 \cdot (- 8)}}{2 \cdot 7}$ $y=(--5+-sqrt((-5)^2-4*7*(-8)))/(2*7)$

$y = \frac{+ 5 \pm \sqrt{25 + 224}}{14}$ $y=(+5+-sqrt(25+224))/(14)$

$y = \frac{+ 5 \pm \sqrt{249}}{14}$ $y=(+5+-sqrt(249))/(14)$

$y_{1} = \frac{+ 5 + \sqrt{249}}{14}$ $y_1=(+5+sqrt(249))/(14)$

$y_{2} = \frac{+ 5 - \sqrt{249}}{14}$ $y_2=(+5-sqrt(249))/(14)$

God bless.... I hope the explanation is useful.

How do you solve 7y2−5y−8=07y^2-5y-8=0 using the quadratic formula?