how do you find the solutions of 2x^2 - 3 = 13?

Look up the Quadratic Formula. Quadratics always have two roots.

1st problem:
2x² - 3 = 13
2x² = 16
x² = 8
x = +/- √8

Answer: x = +/- √8

2nd problem:
x² - 3x = - 2x - 7
x² - x = - 7
x² - 1/2x = - 7 + (- 1/2)²
x² - 1/2x = - 28/4 + 1/4
(x - 1/2)² = - 27/4
x - 1/2 = (√27i)/2

Values of x;
x = (1 + [√27i])/2
x = (1 - [√27i])/2

Answer: x = (1 + [√27i])/2, (1 - [√27i])/2

3rd problem:
x² - 81 = 0
x² = 81
x = +/- 9

Answer: (x + 9)(x - 9) are the factors.

2 x ² = 16
x ² = 8
x = ± √8
x = ± 2√2

x² - x + 7 = 0
x = [ 1 ± √ (1 - 28) ] / 2
x = [ 1 ± √ (- 27) ] / 2
x = [ 1 ± √ ( 27 i ² ) ] / 2
x = [ 1 ± i 3√3 ] / 2

(x - 9)(x + 9) = 0
x = 9 , x = - 9

1)
2x^2 - 3 = 13
2x^2 = 13 + 3
2x^2 = 16
x^2 = 16/2
x^2 = 8
x = ±√8
x = ±√(2^2 * 2)
x = ±2√2

= = = = = = = =

2)
x^2 - 3x = -2x - 7
x^2 - 3x + 2x + 7 = 0
x^2 - x + 7 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = -1
c = 7

x = [1 ±√(1 - 28)]/2
x = [1 ±√-27]/2 (imaginary number)
(no real roots)

= = = = = = = =

3)
x^2 - 81 = 0
x^2 = 81
x = ±√81
x = ±√(9 * 9)
x = ±9

x^2 -3x = -2x - 7
x^2-x+7=0
discriminant=(-1)^2-4(7)
discriminant=-27
thus, no solution


x^2 - 81 = 0
Difference of prefect squares
(x-9)(x+9)=0
x=9 or x=-9

2x^2-3 = 13
2x^2 -3 +3 = 13 +3
2x^2 = 16
2x^2/2 = 16/2
x^2 = 8
x = square root of 8

x^2-3x = -2x-7
x^2 -3x + 2x +7 = 0
x^2 -x + 7 = 0
and it seems like you have to use the quadratic formula find your answer, but i am too lazy and tired to do it now


x^2 - 81= 0
(x+9)(x-9) = 0
x = 9 or -9

2x^2 - 3 = 13
2x^2 = 13+3
2x^2 = 16
x^2 = 16/2
x^2 = 8
x = squareroot of 8
x = 2square root of 2

x^2 - 81 = 0
x^2 = 81
x = squareroot of 81
x = 9

x^2 - 3x = -2x-7
x^2-x+7 = 0
(x+6)(x-7) = 0
x = -6 or x=7

Hope this helps.