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12t^2+4t-5=0
a = 12; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·12·(-5)
Δ = 256
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{256}=16$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*12}=\frac{-20}{24} =-5/6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*12}=\frac{12}{24} =1/2 $
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