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(2x+20)(2x+10)=96
We move all terms to the left:
(2x+20)(2x+10)-(96)=0
We multiply parentheses ..
(+4x^2+20x+40x+200)-96=0
We get rid of parentheses
4x^2+20x+40x+200-96=0
We add all the numbers together, and all the variables
4x^2+60x+104=0
a = 4; b = 60; c = +104;
Δ = b2-4ac
Δ = 602-4·4·104
Δ = 1936
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1936}=44$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-44}{2*4}=\frac{-104}{8} =-13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+44}{2*4}=\frac{-16}{8} =-2 $
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