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5x(2x+4)=40
We move all terms to the left:
5x(2x+4)-(40)=0
We multiply parentheses
10x^2+20x-40=0
a = 10; b = 20; c = -40;
Δ = b2-4ac
Δ = 202-4·10·(-40)
Δ = 2000
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{5}}{2*10}=\frac{-20-20\sqrt{5}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{5}}{2*10}=\frac{-20+20\sqrt{5}}{20} $
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