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-t^2-12t=32
We move all terms to the left:
-t^2-12t-(32)=0
We add all the numbers together, and all the variables
-1t^2-12t-32=0
a = -1; b = -12; c = -32;
Δ = b2-4ac
Δ = -122-4·(-1)·(-32)
Δ = 16
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4}{2*-1}=\frac{8}{-2} =-4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*-1}=\frac{16}{-2} =-8 $
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