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4t^2+18t+8=0
a = 4; b = 18; c = +8;
Δ = b2-4ac
Δ = 182-4·4·8
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-14}{2*4}=\frac{-32}{8} =-4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+14}{2*4}=\frac{-4}{8} =-1/2 $
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