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12t^2+56t+65=0
a = 12; b = 56; c = +65;
Δ = b2-4ac
Δ = 562-4·12·65
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-4}{2*12}=\frac{-60}{24} =-2+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+4}{2*12}=\frac{-52}{24} =-2+1/6 $
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