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4t^2+21t+15=0
a = 4; b = 21; c = +15;
Δ = b2-4ac
Δ = 212-4·4·15
Δ = 201
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{201}}{2*4}=\frac{-21-\sqrt{201}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{201}}{2*4}=\frac{-21+\sqrt{201}}{8} $
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