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-2x(2x-5)=5
We move all terms to the left:
-2x(2x-5)-(5)=0
We multiply parentheses
-4x^2+10x-5=0
a = -4; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·(-4)·(-5)
Δ = 20
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{5}}{2*-4}=\frac{-10-2\sqrt{5}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{5}}{2*-4}=\frac{-10+2\sqrt{5}}{-8} $
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