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3x^2+35x+36=0
a = 3; b = 35; c = +36;
Δ = b2-4ac
Δ = 352-4·3·36
Δ = 793
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-\sqrt{793}}{2*3}=\frac{-35-\sqrt{793}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+\sqrt{793}}{2*3}=\frac{-35+\sqrt{793}}{6} $
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