2x^2-7x+10=40

Solution for 2x^2-7x+10=40 equation:

2x^2-7x+10=40

We move all terms to the left:

2x^2-7x+10-(40)=0

We add all the numbers together, and all the variables

2x^2-7x-30=0

a = 2; b = -7; c = -30;
Δ = b2-4ac
Δ = -72-4·2·(-30)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*2}=\frac{-10}{4}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*2}=\frac{24}{4}
=6
$