# (4x-3)squared=4 solve the equation 4x squared +12x=0

(4x-3)squared=4 solve the equation
4x squared +12x=0
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vibrerentb
For the first one:
$\left(4x-3{\right)}^{2}=4$ Take the square root of both sides, giving you:
4x-3=2
and
4x-3=-2
Solve both for x:
4x-3=2
4x=5
x=5/4
And
4x-3=-2
4x=1
x=1/4
So for the first one you get x=5/4 and x=1/4
For the second one, $4{x}^{2}+12x=0$ start by factoring out anx:
$4{x}^{2}+12x=0$
x(4x+12)=0
From here you can say that x=0 because if you multiplied 0 bywhatever is inside the parenthesis then you would get 0. Then go onand solve for any other x's:
4x+12=0
4x=-12
x=-3
So for the second one you get x=0 and x=-3
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wendi1019gt
when you have a quadratic you have to set it equal to zero and then solve for the zeroes.
$\left(4x-3{\right)}^{2}=4$
Then you foil which multiplies all the values
(4x-3)(4x-3)=4
$16{x}^{2}-12x-12x+9=4$
$16{x}^{2}-24x+9=4$
$16{x}^{2}-24x+5=0$
(4x-5)(4x-1)=0
Then set each part equal to zero and solve for X
4x-5=0
$x=\frac{5}{4}←$ first solution
4x-1=0
$x=\frac{1}{4}←$ second solution
For the second question it already equals zero and because there is no C term we will just factor out an x and then we can solve
$4{x}^{2}+12=0$
x(4x+12)=0
Set each part equal to zero
$x=0←$ 1st solution
4x+12=0
$x=-\frac{12}{4}=-3←$ 2nd solution