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4u(u-3)=6-u
We move all terms to the left:
4u(u-3)-(6-u)=0
We add all the numbers together, and all the variables
4u(u-3)-(-1u+6)=0
We multiply parentheses
4u^2-12u-(-1u+6)=0
We get rid of parentheses
4u^2-12u+1u-6=0
We add all the numbers together, and all the variables
4u^2-11u-6=0
a = 4; b = -11; c = -6;
Δ = b2-4ac
Δ = -112-4·4·(-6)
Δ = 217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{217}}{2*4}=\frac{11-\sqrt{217}}{8} $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{217}}{2*4}=\frac{11+\sqrt{217}}{8} $
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