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t^2+12t-28=0
a = 1; b = 12; c = -28;
Δ = b2-4ac
Δ = 122-4·1·(-28)
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-16}{2*1}=\frac{-28}{2} =-14 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16}{2*1}=\frac{4}{2} =2 $
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