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(7x-16)(3x+4)=180
We move all terms to the left:
(7x-16)(3x+4)-(180)=0
We multiply parentheses ..
(+21x^2+28x-48x-64)-180=0
We get rid of parentheses
21x^2+28x-48x-64-180=0
We add all the numbers together, and all the variables
21x^2-20x-244=0
a = 21; b = -20; c = -244;
Δ = b2-4ac
Δ = -202-4·21·(-244)
Δ = 20896
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20896}=\sqrt{16*1306}=\sqrt{16}*\sqrt{1306}=4\sqrt{1306}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{1306}}{2*21}=\frac{20-4\sqrt{1306}}{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{1306}}{2*21}=\frac{20+4\sqrt{1306}}{42} $
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