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(2t^2-3t-4)=0
We get rid of parentheses
2t^2-3t-4=0
a = 2; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·2·(-4)
Δ = 41
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{41}}{2*2}=\frac{3-\sqrt{41}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{41}}{2*2}=\frac{3+\sqrt{41}}{4} $
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