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5x(2x-4)=25
We move all terms to the left:
5x(2x-4)-(25)=0
We multiply parentheses
10x^2-20x-25=0
a = 10; b = -20; c = -25;
Δ = b2-4ac
Δ = -202-4·10·(-25)
Δ = 1400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1400}=\sqrt{100*14}=\sqrt{100}*\sqrt{14}=10\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{14}}{2*10}=\frac{20-10\sqrt{14}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{14}}{2*10}=\frac{20+10\sqrt{14}}{20} $
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