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8t^2+34t=0
a = 8; b = 34; c = 0;
Δ = b2-4ac
Δ = 342-4·8·0
Δ = 1156
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{1156}=34$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-34}{2*8}=\frac{-68}{16} =-4+1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+34}{2*8}=\frac{0}{16} =0 $
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