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0=-16x^2+22x
We move all terms to the left:
0-(-16x^2+22x)=0
We add all the numbers together, and all the variables
-(-16x^2+22x)=0
We get rid of parentheses
16x^2-22x=0
a = 16; b = -22; c = 0;
Δ = b2-4ac
Δ = -222-4·16·0
Δ = 484
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{484}=22$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-22}{2*16}=\frac{0}{32} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+22}{2*16}=\frac{44}{32} =1+3/8 $
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