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4x^2+32x=25
We move all terms to the left:
4x^2+32x-(25)=0
a = 4; b = 32; c = -25;
Δ = b2-4ac
Δ = 322-4·4·(-25)
Δ = 1424
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1424}=\sqrt{16*89}=\sqrt{16}*\sqrt{89}=4\sqrt{89}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{89}}{2*4}=\frac{-32-4\sqrt{89}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{89}}{2*4}=\frac{-32+4\sqrt{89}}{8} $
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