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(x+63)(4x)=120
We move all terms to the left:
(x+63)(4x)-(120)=0
We multiply parentheses
4x^2+252x-120=0
a = 4; b = 252; c = -120;
Δ = b2-4ac
Δ = 2522-4·4·(-120)
Δ = 65424
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{65424}=\sqrt{16*4089}=\sqrt{16}*\sqrt{4089}=4\sqrt{4089}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(252)-4\sqrt{4089}}{2*4}=\frac{-252-4\sqrt{4089}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(252)+4\sqrt{4089}}{2*4}=\frac{-252+4\sqrt{4089}}{8} $
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