9-2x^2/5=1

Solution for 9-2x^2/5=1 equation:

9-2x^2/5=1

We move all terms to the left:

9-2x^2/5-(1)=0

We add all the numbers together, and all the variables

-2x^2/5+8=0

We multiply all the terms by the denominator


-2x^2+8*5=0

We add all the numbers together, and all the variables

-2x^2+40=0

a = -2; b = 0; c = +40;
Δ = b2-4ac
Δ = 02-4·(-2)·40
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*-2}=\frac{0-8\sqrt{5}}{-4}
=-\frac{8\sqrt{5}}{-4}
=-\frac{2\sqrt{5}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*-2}=\frac{0+8\sqrt{5}}{-4}
=\frac{8\sqrt{5}}{-4}
=\frac{2\sqrt{5}}{-1}
$