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4t^2+5t-250=0
a = 4; b = 5; c = -250;
Δ = b2-4ac
Δ = 52-4·4·(-250)
Δ = 4025
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4025}=\sqrt{25*161}=\sqrt{25}*\sqrt{161}=5\sqrt{161}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{161}}{2*4}=\frac{-5-5\sqrt{161}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{161}}{2*4}=\frac{-5+5\sqrt{161}}{8} $
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