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5x+20=32x^2
We move all terms to the left:
5x+20-(32x^2)=0
determiningTheFunctionDomain -32x^2+5x+20=0
a = -32; b = 5; c = +20;
Δ = b2-4ac
Δ = 52-4·(-32)·20
Δ = 2585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{2585}}{2*-32}=\frac{-5-\sqrt{2585}}{-64} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{2585}}{2*-32}=\frac{-5+\sqrt{2585}}{-64} $
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