How do you use the quadratic formula to find both solutions to the quadratic equation $(2 y - 3) (y + 1) = 5$ $(2y - 3) (y + 1) = 5$ ?

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Answer 1 · 2015-07-06T18:22:14

Multiply out and rearrange into the form $a y^{2} + b y + c = 0$ $ay^2+by+c = 0$ then use the quadratic formula to find:

$y = \frac{1 \pm \sqrt{65}}{4}$ $y = (1+-sqrt(65))/4$

$5 = (2 y - 3) (y + 1) = 2 y^{2} - y - 3$ $5 = (2y-3)(y+1) = 2y^2-y-3$

Subtract $5$ $5$ from both sides to get:

$2 y^{2} - y - 8 = 0$ $2y^2-y-8 = 0$

This is of the form $a y^{2} + b y + c = 0$ $ay^2+by+c = 0$

with $a = 2$ $a=2$ , $b = - 1$ $b=-1$ and $c = - 8$ $c=-8$

This has solutions given by the quadratic formula:

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

$= \frac{1 \pm \sqrt{{(- 1)}^{2} - (4 \times 2 \times - 8)}}{2 \cdot 2}$ $=(1+-sqrt((-1)^2-(4xx2xx-8)))/(2*2)$

$= \frac{1 \pm \sqrt{65}}{4}$ $=(1+-sqrt(65))/4$

How do you use the quadratic formula to find both solutions to the quadratic equation (2y−3)(y+1)=5(2y - 3) (y + 1) = 5?