(a+3)(a-1)-(2a-1)(a+4)=0

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Solution for (a+3)(a-1)-(2a-1)(a+4)=0 equation:



(a+3)(a-1)-(2a-1)(a+4)=0
We multiply parentheses ..
(+a^2-1a+3a-3)-(2a-1)(a+4)=0
We get rid of parentheses
a^2-1a+3a-(2a-1)(a+4)-3=0
We multiply parentheses ..
a^2-(+2a^2+8a-1a-4)-1a+3a-3=0
We add all the numbers together, and all the variables
a^2-(+2a^2+8a-1a-4)+2a-3=0
We get rid of parentheses
a^2-2a^2-8a+1a+2a+4-3=0
We add all the numbers together, and all the variables
-1a^2-5a+1=0
a = -1; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·(-1)·1
Δ = 29
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{29}}{2*-1}=\frac{5-\sqrt{29}}{-2} $
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{29}}{2*-1}=\frac{5+\sqrt{29}}{-2} $

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