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(2x+20)(2x+10)-(10)(20)=216
We move all terms to the left:
(2x+20)(2x+10)-(10)(20)-(216)=0
We add all the numbers together, and all the variables
(2x+20)(2x+10)-1236=0
We multiply parentheses ..
(+4x^2+20x+40x+200)-1236=0
We get rid of parentheses
4x^2+20x+40x+200-1236=0
We add all the numbers together, and all the variables
4x^2+60x-1036=0
a = 4; b = 60; c = -1036;
Δ = b2-4ac
Δ = 602-4·4·(-1036)
Δ = 20176
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20176}=\sqrt{16*1261}=\sqrt{16}*\sqrt{1261}=4\sqrt{1261}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-4\sqrt{1261}}{2*4}=\frac{-60-4\sqrt{1261}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+4\sqrt{1261}}{2*4}=\frac{-60+4\sqrt{1261}}{8} $
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