# How do you solve  -4x^2 +8x - 3 = 0 by completing the square?

Solution: $x = \frac{3}{2} \mathmr{and} x = \frac{1}{2}$
$- 4 {x}^{2} + 8 x - 3 = 0 \mathmr{and} - 4 \left({x}^{2} - 2 x\right) - 3 = 0 \mathmr{and} - 4 \left({x}^{2} - 2 x + 1\right) + 4 - 3 = 0 \mathmr{and} - 4 {\left(x - 1\right)}^{2} + 1 = 0 \mathmr{and} 4 {\left(x - 1\right)}^{2} = 1 \mathmr{and} {\left(x - 1\right)}^{2} = \frac{1}{4} \mathmr{and} x - 1 = \pm \frac{1}{2} \mathmr{and} x = 1 + \frac{1}{2} = \frac{3}{2} \mathmr{and} x = 1 - \frac{1}{2} = \frac{1}{2} \therefore$. Solution: $x = \frac{3}{2} \mathmr{and} x = \frac{1}{2}$[Ans]