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0=5t^2-30t-45
We move all terms to the left:
0-(5t^2-30t-45)=0
We add all the numbers together, and all the variables
-(5t^2-30t-45)=0
We get rid of parentheses
-5t^2+30t+45=0
a = -5; b = 30; c = +45;
Δ = b2-4ac
Δ = 302-4·(-5)·45
Δ = 1800
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-30\sqrt{2}}{2*-5}=\frac{-30-30\sqrt{2}}{-10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+30\sqrt{2}}{2*-5}=\frac{-30+30\sqrt{2}}{-10} $
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