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9t^2+54t-58=0
a = 9; b = 54; c = -58;
Δ = b2-4ac
Δ = 542-4·9·(-58)
Δ = 5004
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5004}=\sqrt{36*139}=\sqrt{36}*\sqrt{139}=6\sqrt{139}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-6\sqrt{139}}{2*9}=\frac{-54-6\sqrt{139}}{18} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+6\sqrt{139}}{2*9}=\frac{-54+6\sqrt{139}}{18} $
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