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10x(4x+4)=400
We move all terms to the left:
10x(4x+4)-(400)=0
We multiply parentheses
40x^2+40x-400=0
a = 40; b = 40; c = -400;
Δ = b2-4ac
Δ = 402-4·40·(-400)
Δ = 65600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{65600}=\sqrt{1600*41}=\sqrt{1600}*\sqrt{41}=40\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40\sqrt{41}}{2*40}=\frac{-40-40\sqrt{41}}{80} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40\sqrt{41}}{2*40}=\frac{-40+40\sqrt{41}}{80} $
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