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2x^2-25x=225
We move all terms to the left:
2x^2-25x-(225)=0
a = 2; b = -25; c = -225;
Δ = b2-4ac
Δ = -252-4·2·(-225)
Δ = 2425
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2425}=\sqrt{25*97}=\sqrt{25}*\sqrt{97}=5\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{97}}{2*2}=\frac{25-5\sqrt{97}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{97}}{2*2}=\frac{25+5\sqrt{97}}{4} $
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