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2x^2-10x+16=8
We move all terms to the left:
2x^2-10x+16-(8)=0
We add all the numbers together, and all the variables
2x^2-10x+8=0
a = 2; b = -10; c = +8;
Δ = b2-4ac
Δ = -102-4·2·8
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6}{2*2}=\frac{4}{4} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6}{2*2}=\frac{16}{4} =4 $
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