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16t^2+28t-25=0
a = 16; b = 28; c = -25;
Δ = b2-4ac
Δ = 282-4·16·(-25)
Δ = 2384
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2384}=\sqrt{16*149}=\sqrt{16}*\sqrt{149}=4\sqrt{149}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{149}}{2*16}=\frac{-28-4\sqrt{149}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{149}}{2*16}=\frac{-28+4\sqrt{149}}{32} $
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