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2t^2-11t-30=0
a = 2; b = -11; c = -30;
Δ = b2-4ac
Δ = -112-4·2·(-30)
Δ = 361
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{361}=19$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2*2}=\frac{-8}{4} =-2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*2}=\frac{30}{4} =7+1/2 $
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