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2t^2+15t+22=0
a = 2; b = 15; c = +22;
Δ = b2-4ac
Δ = 152-4·2·22
Δ = 49
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{49}=7$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-7}{2*2}=\frac{-22}{4} =-5+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+7}{2*2}=\frac{-8}{4} =-2 $
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