(4x^2)+(3x^2)=25

Solution for (4x^2)+(3x^2)=25 equation:

(4x^2)+(3x^2)=25

We move all terms to the left:

(4x^2)+(3x^2)-(25)=0

We add all the numbers together, and all the variables

7x^2-25=0

a = 7; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·7·(-25)
Δ = 700
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:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{7}}{2*7}=\frac{0-10\sqrt{7}}{14}
=-\frac{10\sqrt{7}}{14}
=-\frac{5\sqrt{7}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{7}}{2*7}=\frac{0+10\sqrt{7}}{14}
=\frac{10\sqrt{7}}{14}
=\frac{5\sqrt{7}}{7}
$