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125=5t+10t^2
We move all terms to the left:
125-(5t+10t^2)=0
We get rid of parentheses
-10t^2-5t+125=0
a = -10; b = -5; c = +125;
Δ = b2-4ac
Δ = -52-4·(-10)·125
Δ = 5025
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5025}=\sqrt{25*201}=\sqrt{25}*\sqrt{201}=5\sqrt{201}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{201}}{2*-10}=\frac{5-5\sqrt{201}}{-20} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{201}}{2*-10}=\frac{5+5\sqrt{201}}{-20} $
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