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(4x-20)(3x+14)=155
We move all terms to the left:
(4x-20)(3x+14)-(155)=0
We multiply parentheses ..
(+12x^2+56x-60x-280)-155=0
We get rid of parentheses
12x^2+56x-60x-280-155=0
We add all the numbers together, and all the variables
12x^2-4x-435=0
a = 12; b = -4; c = -435;
Δ = b2-4ac
Δ = -42-4·12·(-435)
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{-(-4)-4\sqrt{1306}}{2*12}=\frac{4-4\sqrt{1306}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{1306}}{2*12}=\frac{4+4\sqrt{1306}}{24} $
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