4y^2+7y^2-36=0

Solution for 4y^2+7y^2-36=0 equation:

4y^2+7y^2-36=0

We add all the numbers together, and all the variables

11y^2-36=0

a = 11; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·11·(-36)
Δ = 1584
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1584}=\sqrt{144*11}=\sqrt{144}*\sqrt{11}=12\sqrt{11}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{11}}{2*11}=\frac{0-12\sqrt{11}}{22}
=-\frac{12\sqrt{11}}{22}
=-\frac{6\sqrt{11}}{11}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{11}}{2*11}=\frac{0+12\sqrt{11}}{22}
=\frac{12\sqrt{11}}{22}
=\frac{6\sqrt{11}}{11}
$