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23t^2+48t=0
a = 23; b = 48; c = 0;
Δ = b2-4ac
Δ = 482-4·23·0
Δ = 2304
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{2304}=48$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48}{2*23}=\frac{-96}{46} =-2+2/23 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48}{2*23}=\frac{0}{46} =0 $
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