How do you solve 5x^2-3x+3=-2x^2+3?

Jan 23, 2017

$x = 0 \mathmr{and} x = + \frac{3}{7}$

Explanation:

Given:$\text{ } 5 {x}^{2} - 3 x + 3 = - 2 {x}^{2} + 3$

Spot that we have:$\text{ } 5 {x}^{2} - 3 x + \cancel{3} = - 2 {x}^{2} + \cancel{3}$

Add $2 {x}^{2}$ to both sides

$5 {x}^{2} + 2 {x}^{2} - 3 x = - 2 {x}^{2} + 2 {x}^{2}$

$7 {x}^{2} - 3 x = 0$

Factor out the $x$

$x \left(7 x - 3\right) = 0$

$x = 0$ is 1 solution

The other solution is that $7 x - 3 = 0$

$\implies 7 x = + 3$

$\implies x = + \frac{3}{7}$
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$\textcolor{b l u e}{\text{Plotting } y = 5 {x}^{2} - 3 x + 3 \mathmr{and}}$
$\textcolor{b l u e}{y = - 2 {x}^{2} + 3 \text{ gives the same result}}$