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(-16t^2)+30t+6=0
We get rid of parentheses
-16t^2+30t+6=0
a = -16; b = 30; c = +6;
Δ = b2-4ac
Δ = 302-4·(-16)·6
Δ = 1284
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{321}}{2*-16}=\frac{-30-2\sqrt{321}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{321}}{2*-16}=\frac{-30+2\sqrt{321}}{-32} $
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