(7+y)(4y+3)=0

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Solution for (7+y)(4y+3)=0 equation:



(7+y)(4y+3)=0
We add all the numbers together, and all the variables
(y+7)(4y+3)=0
We multiply parentheses ..
(+4y^2+3y+28y+21)=0
We get rid of parentheses
4y^2+3y+28y+21=0
We add all the numbers together, and all the variables
4y^2+31y+21=0
a = 4; b = 31; c = +21;
Δ = b2-4ac
Δ = 312-4·4·21
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-25}{2*4}=\frac{-56}{8} =-7 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+25}{2*4}=\frac{-6}{8} =-3/4 $

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