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12=36t-16t^2
We move all terms to the left:
12-(36t-16t^2)=0
We get rid of parentheses
16t^2-36t+12=0
a = 16; b = -36; c = +12;
Δ = b2-4ac
Δ = -362-4·16·12
Δ = 528
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{33}}{2*16}=\frac{36-4\sqrt{33}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{33}}{2*16}=\frac{36+4\sqrt{33}}{32} $
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