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2x(x-5)=36
We move all terms to the left:
2x(x-5)-(36)=0
We multiply parentheses
2x^2-10x-36=0
a = 2; b = -10; c = -36;
Δ = b2-4ac
Δ = -102-4·2·(-36)
Δ = 388
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{97}}{2*2}=\frac{10-2\sqrt{97}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{97}}{2*2}=\frac{10+2\sqrt{97}}{4} $
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