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(4x-6)(x-4)=120
We move all terms to the left:
(4x-6)(x-4)-(120)=0
We multiply parentheses ..
(+4x^2-16x-6x+24)-120=0
We get rid of parentheses
4x^2-16x-6x+24-120=0
We add all the numbers together, and all the variables
4x^2-22x-96=0
a = 4; b = -22; c = -96;
Δ = b2-4ac
Δ = -222-4·4·(-96)
Δ = 2020
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{505}}{2*4}=\frac{22-2\sqrt{505}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{505}}{2*4}=\frac{22+2\sqrt{505}}{8} $
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