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5t-10t^2=-10
We move all terms to the left:
5t-10t^2-(-10)=0
We add all the numbers together, and all the variables
-10t^2+5t+10=0
a = -10; b = 5; c = +10;
Δ = b2-4ac
Δ = 52-4·(-10)·10
Δ = 425
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{17}}{2*-10}=\frac{-5-5\sqrt{17}}{-20} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{17}}{2*-10}=\frac{-5+5\sqrt{17}}{-20} $
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