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4x^2+5x=25
We move all terms to the left:
4x^2+5x-(25)=0
a = 4; b = 5; c = -25;
Δ = b2-4ac
Δ = 52-4·4·(-25)
Δ = 425
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{17}}{2*4}=\frac{-5-5\sqrt{17}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{17}}{2*4}=\frac{-5+5\sqrt{17}}{8} $
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