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4x^2+92x-692=0
a = 4; b = 92; c = -692;
Δ = b2-4ac
Δ = 922-4·4·(-692)
Δ = 19536
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{19536}=\sqrt{16*1221}=\sqrt{16}*\sqrt{1221}=4\sqrt{1221}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(92)-4\sqrt{1221}}{2*4}=\frac{-92-4\sqrt{1221}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(92)+4\sqrt{1221}}{2*4}=\frac{-92+4\sqrt{1221}}{8} $
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