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X^2+3=10X-6
We move all terms to the left:
X^2+3-(10X-6)=0
We get rid of parentheses
X^2-10X+6+3=0
We add all the numbers together, and all the variables
X^2-10X+9=0
a = 1; b = -10; c = +9;
Δ = b2-4ac
Δ = -102-4·1·9
Δ = 64
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{64}=8$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-8}{2*1}=\frac{2}{2} =1 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+8}{2*1}=\frac{18}{2} =9 $
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