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5x(2x-13)=40
We move all terms to the left:
5x(2x-13)-(40)=0
We multiply parentheses
10x^2-65x-40=0
a = 10; b = -65; c = -40;
Δ = b2-4ac
Δ = -652-4·10·(-40)
Δ = 5825
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5825}=\sqrt{25*233}=\sqrt{25}*\sqrt{233}=5\sqrt{233}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-5\sqrt{233}}{2*10}=\frac{65-5\sqrt{233}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+5\sqrt{233}}{2*10}=\frac{65+5\sqrt{233}}{20} $
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