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2t^2+30=16t
We move all terms to the left:
2t^2+30-(16t)=0
a = 2; b = -16; c = +30;
Δ = b2-4ac
Δ = -162-4·2·30
Δ = 16
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{16}=4$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*2}=\frac{12}{4} =3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*2}=\frac{20}{4} =5 $
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