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4t^2+27t+18=0
a = 4; b = 27; c = +18;
Δ = b2-4ac
Δ = 272-4·4·18
Δ = 441
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{441}=21$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-21}{2*4}=\frac{-48}{8} =-6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+21}{2*4}=\frac{-6}{8} =-3/4 $
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