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4x^2+36x-213=0
a = 4; b = 36; c = -213;
Δ = b2-4ac
Δ = 362-4·4·(-213)
Δ = 4704
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4704}=\sqrt{784*6}=\sqrt{784}*\sqrt{6}=28\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-28\sqrt{6}}{2*4}=\frac{-36-28\sqrt{6}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+28\sqrt{6}}{2*4}=\frac{-36+28\sqrt{6}}{8} $
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