What are the solutions to the equation bx^2+5=6 ?

The solutions to the equation bx^2+5=6 are x= 1/(sqrt(b)),x=-1/(sqrt(b));\quad b\ne 0

Find the zeros of bx^2+5=6

The zeros of bx^2+5=6 are x= 1/(sqrt(b)),x=-1/(sqrt(b));\quad b\ne 0

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b x^{2} + 5 = 6

x = \frac{1}{b}, x = - \frac{1}{b}; b \neq = 0

Solution steps

b x^{2} + 5 = 6

Move

5

to the right side

b x^{2} = 1

Divide both sides by

b; b \neq = 0

x^{2} = \frac{1}{b}; b \neq = 0

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = \frac{1}{b}, x = - \frac{1}{b}; b \neq = 0

Simplify

\frac{1}{b} : \frac{1}{b}

Simplify

- \frac{1}{b} : - \frac{1}{b}

x = \frac{1}{b}, x = - \frac{1}{b}; b \neq = 0

3 0^{2} + 4 0^{2} = c^{2}

3 x^{2} - x + 15 = 0

x^{2} + 88 = 10 x

5 x^{2} + 20 x + 2 = 0

8 c^{2} + 2 c - 1 = 0

What are the solutions to the equation bx^2+5=6 ?
The solutions to the equation bx^2+5=6 are x= 1/(sqrt(b)),x=-1/(sqrt(b));\quad b\ne 0
Find the zeros of bx^2+5=6
The zeros of bx^2+5=6 are x= 1/(sqrt(b)),x=-1/(sqrt(b));\quad b\ne 0