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5t^2+3t-92=0
a = 5; b = 3; c = -92;
Δ = b2-4ac
Δ = 32-4·5·(-92)
Δ = 1849
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{1849}=43$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-43}{2*5}=\frac{-46}{10} =-4+3/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+43}{2*5}=\frac{40}{10} =4 $
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