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(12x*12x)-(3x-4)(4x+1)=19
We move all terms to the left:
(12x*12x)-(3x-4)(4x+1)-(19)=0
We add all the numbers together, and all the variables
(+12x*12x)-(3x-4)(4x+1)-19=0
We get rid of parentheses
12x*12x-(3x-4)(4x+1)-19=0
We multiply parentheses ..
-(+12x^2+3x-16x-4)+12x*12x-19=0
Wy multiply elements
-(+12x^2+3x-16x-4)+144x^2-19=0
We get rid of parentheses
-12x^2+144x^2-3x+16x+4-19=0
We add all the numbers together, and all the variables
132x^2+13x-15=0
a = 132; b = 13; c = -15;
Δ = b2-4ac
Δ = 132-4·132·(-15)
Δ = 8089
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{8089}}{2*132}=\frac{-13-\sqrt{8089}}{264} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{8089}}{2*132}=\frac{-13+\sqrt{8089}}{264} $
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