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4t^2+15t-1=0
a = 4; b = 15; c = -1;
Δ = b2-4ac
Δ = 152-4·4·(-1)
Δ = 241
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{-(15)-\sqrt{241}}{2*4}=\frac{-15-\sqrt{241}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{241}}{2*4}=\frac{-15+\sqrt{241}}{8} $
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