(2x^2-15)=(x^2+34)

Solution for (2x^2-15)=(x^2+34) equation:

(2x^2-15)=(x^2+34)

We move all terms to the left:

(2x^2-15)-((x^2+34))=0

We get rid of parentheses

2x^2-((x^2+34))-15=0


We calculate terms in parentheses: -((x^2+34)), so:

(x^2+34)

We get rid of parentheses

x^2+34

Back to the equation:

-(x^2+34)


We get rid of parentheses

2x^2-x^2-34-15=0

We add all the numbers together, and all the variables

x^2-49=0

a = 1; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·1·(-49)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14}{2*1}=\frac{-14}{2}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14}{2*1}=\frac{14}{2}
=7
$