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2x(x+10)=40
We move all terms to the left:
2x(x+10)-(40)=0
We multiply parentheses
2x^2+20x-40=0
a = 2; b = 20; c = -40;
Δ = b2-4ac
Δ = 202-4·2·(-40)
Δ = 720
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{5}}{2*2}=\frac{-20-12\sqrt{5}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{5}}{2*2}=\frac{-20+12\sqrt{5}}{4} $
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