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4x^2=16x=
We move all terms to the left:
4x^2-(16x)=0
a = 4; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·4·0
Δ = 256
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*4}=\frac{0}{8} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*4}=\frac{32}{8} =4 $
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