How do you find the roots, real and imaginary, of $y = (4 x - 5) (x + 1)$ $y=(4x-5)(x+1)$ using the quadratic formula?

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Answer 1 · 2018-05-16T19:38:35

$x_{1} = \frac{5}{4}$ $x_1=5/4$
$x_{2} = - 1$ $x_2=-1$

$(4 x - 5) \cdot (x + 1)$ $(4x-5)*(x+1)$

$= 4 x^{2} + 4 x \cdot 1 - 5 \cdot x - 5$ $=4x^2+4x*1-5*x-5$

$= 4 x^{2} + 4 x - 5 x - 5$ $=4x^2+4x-5x-5$

$= 4 x^{2} - x - 5 =$ $=4x^2-x-5=$

$x_{1, 2} = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (4 (- 5))}}{2 \cdot 4}$ $x_(1,2)=(-(-1)+-sqrt((-1)^2-4(4(-5))))/(2*4)$

$x_{1, 2} = \frac{1 \pm \sqrt{1 + 80}}{8}$ $x_(1,2)=(1+-sqrt(1+80))/8$

$x_{1, 2} = \frac{1 \pm 9}{8}$ $x_(1,2)=(1+-9)/8$

$x_{1} = \frac{5}{4}$ $x_1=5/4$

$x_{2} = - 1$ $x_2=-1$

How do you find the roots, real and imaginary, of y=(4x-5)(x+1)y=(4x−5)(x+1)y=(4x-5)(x+1) using the quadratic formula?