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5x(2x-7)=11
We move all terms to the left:
5x(2x-7)-(11)=0
We multiply parentheses
10x^2-35x-11=0
a = 10; b = -35; c = -11;
Δ = b2-4ac
Δ = -352-4·10·(-11)
Δ = 1665
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1665}=\sqrt{9*185}=\sqrt{9}*\sqrt{185}=3\sqrt{185}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-3\sqrt{185}}{2*10}=\frac{35-3\sqrt{185}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+3\sqrt{185}}{2*10}=\frac{35+3\sqrt{185}}{20} $
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