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3x^2-23x-36=0
a = 3; b = -23; c = -36;
Δ = b2-4ac
Δ = -232-4·3·(-36)
Δ = 961
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-31}{2*3}=\frac{-8}{6} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+31}{2*3}=\frac{54}{6} =9 $
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