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2x^2+22x+36=0
a = 2; b = 22; c = +36;
Δ = b2-4ac
Δ = 222-4·2·36
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-14}{2*2}=\frac{-36}{4} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*2}=\frac{-8}{4} =-2 $
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