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x(2x-5)=10
We move all terms to the left:
x(2x-5)-(10)=0
We multiply parentheses
2x^2-5x-10=0
a = 2; b = -5; c = -10;
Δ = b2-4ac
Δ = -52-4·2·(-10)
Δ = 105
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{105}}{2*2}=\frac{5-\sqrt{105}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{105}}{2*2}=\frac{5+\sqrt{105}}{4} $
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