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4t^2+23t-30=0
a = 4; b = 23; c = -30;
Δ = b2-4ac
Δ = 232-4·4·(-30)
Δ = 1009
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{-(23)-\sqrt{1009}}{2*4}=\frac{-23-\sqrt{1009}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{1009}}{2*4}=\frac{-23+\sqrt{1009}}{8} $
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