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5x(4x+20)=40
We move all terms to the left:
5x(4x+20)-(40)=0
We multiply parentheses
20x^2+100x-40=0
a = 20; b = 100; c = -40;
Δ = b2-4ac
Δ = 1002-4·20·(-40)
Δ = 13200
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{13200}=\sqrt{400*33}=\sqrt{400}*\sqrt{33}=20\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-20\sqrt{33}}{2*20}=\frac{-100-20\sqrt{33}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+20\sqrt{33}}{2*20}=\frac{-100+20\sqrt{33}}{40} $
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