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0=5t^2+15t-10
We move all terms to the left:
0-(5t^2+15t-10)=0
We add all the numbers together, and all the variables
-(5t^2+15t-10)=0
We get rid of parentheses
-5t^2-15t+10=0
a = -5; b = -15; c = +10;
Δ = b2-4ac
Δ = -152-4·(-5)·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{-(-15)-5\sqrt{17}}{2*-5}=\frac{15-5\sqrt{17}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5\sqrt{17}}{2*-5}=\frac{15+5\sqrt{17}}{-10} $
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