10x^2+x-2=3-4x^2

Solution for 10x^2+x-2=3-4x^2 equation:

10x^2+x-2=3-4x^2

We move all terms to the left:

10x^2+x-2-(3-4x^2)=0

We get rid of parentheses

10x^2+4x^2+x-3-2=0

We add all the numbers together, and all the variables

14x^2+x-5=0

a = 14; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·14·(-5)
Δ = 281
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{-(1)-\sqrt{281}}{2*14}=\frac{-1-\sqrt{281}}{28}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{281}}{2*14}=\frac{-1+\sqrt{281}}{28}
$