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4t^2-3t-5=3.2
We move all terms to the left:
4t^2-3t-5-(3.2)=0
We add all the numbers together, and all the variables
4t^2-3t-8.2=0
a = 4; b = -3; c = -8.2;
Δ = b2-4ac
Δ = -32-4·4·(-8.2)
Δ = 140.2
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{140.2}}{2*4}=\frac{3-\sqrt{140.2}}{8} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{140.2}}{2*4}=\frac{3+\sqrt{140.2}}{8} $
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