(5x+2)(x+1)-8=0

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Solution for (5x+2)(x+1)-8=0 equation:



(5x+2)(x+1)-8=0
We multiply parentheses ..
(+5x^2+5x+2x+2)-8=0
We get rid of parentheses
5x^2+5x+2x+2-8=0
We add all the numbers together, and all the variables
5x^2+7x-6=0
a = 5; b = 7; c = -6;
Δ = b2-4ac
Δ = 72-4·5·(-6)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*5}=\frac{-20}{10} =-2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*5}=\frac{6}{10} =3/5 $

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