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(2x+10)(x+5)=234
We move all terms to the left:
(2x+10)(x+5)-(234)=0
We multiply parentheses ..
(+2x^2+10x+10x+50)-234=0
We get rid of parentheses
2x^2+10x+10x+50-234=0
We add all the numbers together, and all the variables
2x^2+20x-184=0
a = 2; b = 20; c = -184;
Δ = b2-4ac
Δ = 202-4·2·(-184)
Δ = 1872
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1872}=\sqrt{144*13}=\sqrt{144}*\sqrt{13}=12\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{13}}{2*2}=\frac{-20-12\sqrt{13}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{13}}{2*2}=\frac{-20+12\sqrt{13}}{4} $
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