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4x^2+33x+36=0
a = 4; b = 33; c = +36;
Δ = b2-4ac
Δ = 332-4·4·36
Δ = 513
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{57}}{2*4}=\frac{-33-3\sqrt{57}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{57}}{2*4}=\frac{-33+3\sqrt{57}}{8} $
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