1400=1250(1+(x)(2))

Solution for 1400=1250(1+(x)(2)) equation:

1400=1250(1+(x)(2))

We move all terms to the left:

1400-(1250(1+(x)(2)))=0

We add all the numbers together, and all the variables

-(1250(+x^2+1))+1400=0


We calculate terms in parentheses: -(1250(+x^2+1)), so:

1250(+x^2+1)

We multiply parentheses

1250x^2+1250

Back to the equation:

-(1250x^2+1250)


We get rid of parentheses

-1250x^2-1250+1400=0

We add all the numbers together, and all the variables

-1250x^2+150=0

a = -1250; b = 0; c = +150;
Δ = b2-4ac
Δ = 02-4·(-1250)·150
Δ = 750000
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:

$\sqrt{\Delta}=\sqrt{750000}=\sqrt{250000*3}=\sqrt{250000}*\sqrt{3}=500\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-500\sqrt{3}}{2*-1250}=\frac{0-500\sqrt{3}}{-2500}
=-\frac{500\sqrt{3}}{-2500}
=-\frac{\sqrt{3}}{-5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+500\sqrt{3}}{2*-1250}=\frac{0+500\sqrt{3}}{-2500}
=\frac{500\sqrt{3}}{-2500}
=\frac{\sqrt{3}}{-5}
$