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8x(5x+5)=13
We move all terms to the left:
8x(5x+5)-(13)=0
We multiply parentheses
40x^2+40x-13=0
a = 40; b = 40; c = -13;
Δ = b2-4ac
Δ = 402-4·40·(-13)
Δ = 3680
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3680}=\sqrt{16*230}=\sqrt{16}*\sqrt{230}=4\sqrt{230}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{230}}{2*40}=\frac{-40-4\sqrt{230}}{80} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{230}}{2*40}=\frac{-40+4\sqrt{230}}{80} $
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