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4b^2+16b-84=0
a = 4; b = 16; c = -84;
Δ = b2-4ac
Δ = 162-4·4·(-84)
Δ = 1600
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{1600}=40$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-40}{2*4}=\frac{-56}{8} =-7 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+40}{2*4}=\frac{24}{8} =3 $
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