How do you solve $4x ^ { 2} - 4x + 2= x ^ { 2} + 6x - 6$ ?

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Answer 1 · 2017-01-30T19:01:33

$x=4/3 = 1 1/3$ or $x=2$

$4x^2-4x+2=x^2+6x-6$

$:.4x^2-x^2-4x-6x=-6-2$

$:.3x^2-10x=-8$

$:.3x^2-10x+8=0$

$:.(3x-4)(x-2=0)$

$:.3x=4 , x=2$

$:.x=4/3 , x=2$

substitute $x= 4/3$

$:.4(4/3)^2-4(4/3)+2=(4/3)^2+6(4/3)-6)$

$:.4(16/9)-16/3+2=16/9+24/3-6$

$:.64/9-16/3+2/1=16/9+24/3-6/1$

$:.(64-48+18=16+72-54)/9$

$:.34=34$

substitute $x= 2$

$:.4(2)^2-4(2)+2=(2)^2+6(2)-6$

$:.16-8+2=4+12-6$

$:.10=10$

How do you solve 4x ^ { 2} - 4x + 2= x ^ { 2} + 6x - 64x ^ { 2} - 4x + 2= x ^ { 2} + 6x - 6?