suchonosdy

2022-07-26

solve by completing the square.
$2{x}^{2}+2x-3=0$

cindysnifflesuz

Expert

$2{x}^{2}+2x-3=0$
Divide through bywhatever is multiplied on the squared term.
${x}^{2}+x-\frac{3}{2}=0$
Take half of thecoefficient (don't forget the sign!) of the x-term, and square it. Add thissquare to both sides of the equation.
${x}^{2}+x-\frac{3}{2}+.25=.25$
Bring the 3/2 over toother side
${x}^{2}+x+.25=\frac{3}{2}+.25$
$\left(x+5{\right)}^{2}=1.75$
square root both sides
$x±.5=\sqrt{1.75}$
$x=\sqrt{1.75}±.5$
$x=\sqrt{\frac{7}{4}}±\frac{1}{2}$
and this can be rewrote:
$x=\frac{\sqrt{7}}{2}±\frac{1}{2}$
$x=\frac{1±\sqrt{7}}{2}$

Alonzo Odom

Expert

I would use the quadratic formula which is:$\left(-b±\sqrt{{b}^{2}-4ac}\right)/2a$
Using that:
$\left(-2±\sqrt{22-4\left(2\right)\left(-3\right)}\right)/2\left(2\right)=\left(-2±\sqrt{28}\right)/4$

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