4x^2+7x=280^2

Solution for 4x^2+7x=280^2 equation:

4x^2+7x=280^2

We move all terms to the left:

4x^2+7x-(280^2)=0

We add all the numbers together, and all the variables

4x^2+7x-78400=0

a = 4; b = 7; c = -78400;
Δ = b2-4ac
Δ = 72-4·4·(-78400)
Δ = 1254449
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1254449}=\sqrt{49*25601}=\sqrt{49}*\sqrt{25601}=7\sqrt{25601}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7\sqrt{25601}}{2*4}=\frac{-7-7\sqrt{25601}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7\sqrt{25601}}{2*4}=\frac{-7+7\sqrt{25601}}{8}
$