How do you solve $3x^2-8x-19=(x-1)^2$ ?

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Answer 1 · 2017-01-26T06:06:50

$x=5, -2$

$3x^2-8x-19=(x-1)^2$

$3x^2-8x-19=x^2-2x+1$

$3x^2color(red)(-x^2)-8xcolor(red)(+2x)-19color(red)(-1)=x^2color(red)(-x^2)-2xcolor(red)(+2x)+1color(red)(-1)$

$2x^2-6x-20=0$

Let's use the quadratic formula to solve for $x$ :

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(6+-sqrt((-6)^2-4(2)(-20)))/(2(2))$

$x=(6+-sqrt(36+160))/4$

$x=(6+-sqrt(196))/4$

$x=(6+-14)/4$

$x=20/4=5, =-8/4=-2$

And we can check this by graphing both sides of the original equation (note the points of intersection at $x=5, -2$ :

graph{(y-3x^2+8x+19)(y-(x-1)^2)=0 [-10, 10, -2.11, 26.78]}

How do you solve 3x^2-8x-19=(x-1)^23x^2-8x-19=(x-1)^2?