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4s^2+16s-9=0
a = 4; b = 16; c = -9;
Δ = b2-4ac
Δ = 162-4·4·(-9)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-20}{2*4}=\frac{-36}{8} =-4+1/2 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+20}{2*4}=\frac{4}{8} =1/2 $
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