3x^2+x=1.25

Solution for 3x^2+x=1.25 equation:

3x^2+x=1.25

We move all terms to the left:

3x^2+x-(1.25)=0

We add all the numbers together, and all the variables

3x^2+x-1.25=0

a = 3; b = 1; c = -1.25;
Δ = b2-4ac
Δ = 12-4·3·(-1.25)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-4}{2*3}=\frac{-5}{6}
=-5/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+4}{2*3}=\frac{3}{6}
=1/2
$