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3x^2+22x-56=0
a = 3; b = 22; c = -56;
Δ = b2-4ac
Δ = 222-4·3·(-56)
Δ = 1156
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{1156}=34$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-34}{2*3}=\frac{-56}{6} =-9+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+34}{2*3}=\frac{12}{6} =2 $
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