How do you use the quadratic formula to solve $\frac{2}{3} x^{2} - 2 = \frac{1}{2} x + 1$ $2/3x^2-2=1/2x+1$ ?

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Answer 1 · 2017-07-25T13:00:58

$x = 1.921$ $x =1.921$

$x = - 1.171$ $x = -1.171$

$\frac{2}{3} x^{2} - 2 = \frac{1}{2} x + 1 \leftarrow$ $2/3x^2-2=1/2x+1" "larr$ get rid of the fractions: $\times 6$ $xx6$

$(cancel6^2xx2)/cancel3x^2-6xx2=(cancel6^3xx1)/cancel2x+6xx1$

$4x^2-12=3x+6" "larr$ make it equal to $0$

$4x^2-3x-18 =0" "larr$ does not factorise

Use the quadratic formula to solve for $x$

$x =(-b+-sqrt(b^2-4ac))/(2a)" "$ with $a=4, b=-3 and c=-18$

$x =(-(-3)+-sqrt((-3)^2-4(4)(-18)))/(2(4))$

$x =(3+-sqrt((9+144)))/(8)$

$x =(3+-sqrt153)/8$

Solve twice to find $x$

$x =1.921$

$x = -1.171$

How do you use the quadratic formula to solve 2/3x^2-2=1/2x+123x2−2=12x+12/3x^2-2=1/2x+1?