x^2+3-7=-2x^2-5x+10

Solution for x^2+3-7=-2x^2-5x+10 equation:

x^2+3-7=-2x^2-5x+10

We move all terms to the left:

x^2+3-7-(-2x^2-5x+10)=0

We add all the numbers together, and all the variables

x^2-(-2x^2-5x+10)-4=0

We get rid of parentheses

x^2+2x^2+5x-10-4=0

We add all the numbers together, and all the variables

3x^2+5x-14=0

a = 3; b = 5; c = -14;
Δ = b2-4ac
Δ = 52-4·3·(-14)
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{193}}{2*3}=\frac{-5-\sqrt{193}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{193}}{2*3}=\frac{-5+\sqrt{193}}{6}
$