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t^2+3t-238=0
a = 1; b = 3; c = -238;
Δ = b2-4ac
Δ = 32-4·1·(-238)
Δ = 961
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{961}=31$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-31}{2*1}=\frac{-34}{2} =-17 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+31}{2*1}=\frac{28}{2} =14 $
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