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3b^2+22b+35=0
a = 3; b = 22; c = +35;
Δ = b2-4ac
Δ = 222-4·3·35
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-8}{2*3}=\frac{-30}{6} =-5 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+8}{2*3}=\frac{-14}{6} =-2+1/3 $
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