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(4x-4)(3x+5)=100
We move all terms to the left:
(4x-4)(3x+5)-(100)=0
We multiply parentheses ..
(+12x^2+20x-12x-20)-100=0
We get rid of parentheses
12x^2+20x-12x-20-100=0
We add all the numbers together, and all the variables
12x^2+8x-120=0
a = 12; b = 8; c = -120;
Δ = b2-4ac
Δ = 82-4·12·(-120)
Δ = 5824
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5824}=\sqrt{64*91}=\sqrt{64}*\sqrt{91}=8\sqrt{91}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{91}}{2*12}=\frac{-8-8\sqrt{91}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{91}}{2*12}=\frac{-8+8\sqrt{91}}{24} $
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