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(4x^2)+36x-242=0
a = 4; b = 36; c = -242;
Δ = b2-4ac
Δ = 362-4·4·(-242)
Δ = 5168
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5168}=\sqrt{16*323}=\sqrt{16}*\sqrt{323}=4\sqrt{323}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{323}}{2*4}=\frac{-36-4\sqrt{323}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{323}}{2*4}=\frac{-36+4\sqrt{323}}{8} $
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