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32-10t=t2
We move all terms to the left:
32-10t-(t2)=0
We add all the numbers together, and all the variables
-1t^2-10t+32=0
a = -1; b = -10; c = +32;
Δ = b2-4ac
Δ = -102-4·(-1)·32
Δ = 228
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{57}}{2*-1}=\frac{10-2\sqrt{57}}{-2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{57}}{2*-1}=\frac{10+2\sqrt{57}}{-2} $
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