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4x^2-5x=16
We move all terms to the left:
4x^2-5x-(16)=0
a = 4; b = -5; c = -16;
Δ = b2-4ac
Δ = -52-4·4·(-16)
Δ = 281
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{281}}{2*4}=\frac{5-\sqrt{281}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{281}}{2*4}=\frac{5+\sqrt{281}}{8} $
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