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4x(x+2)=40
We move all terms to the left:
4x(x+2)-(40)=0
We multiply parentheses
4x^2+8x-40=0
a = 4; b = 8; c = -40;
Δ = b2-4ac
Δ = 82-4·4·(-40)
Δ = 704
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{11}}{2*4}=\frac{-8-8\sqrt{11}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{11}}{2*4}=\frac{-8+8\sqrt{11}}{8} $
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