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4t^2=32t
We move all terms to the left:
4t^2-(32t)=0
a = 4; b = -32; c = 0;
Δ = b2-4ac
Δ = -322-4·4·0
Δ = 1024
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{1024}=32$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-32}{2*4}=\frac{0}{8} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+32}{2*4}=\frac{64}{8} =8 $
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