What are the solutions to the equation b^2-10b+10=-6 ?

The solutions to the equation b^2-10b+10=-6 are b=8,b=2

Find the zeros of b^2-10b+10=-6

The zeros of b^2-10b+10=-6 are b=8,b=2

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b^{2} - 10 b + 10 = - 6

b = 8, b = 2

Solution steps

b^{2} - 10 b + 10 = - 6

Move

6

to the left side

b^{2} - 10 b + 16 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 16 = 6

b_{1, 2} = \frac{- ( - 10 ) \pm 6}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 10 ) + 6}{2 \cdot 1}, b_{2} = \frac{- ( - 10 ) - 6}{2 \cdot 1}

b = \frac{- ( - 10 ) + 6}{2 \cdot 1} : 8

b = \frac{- ( - 10 ) - 6}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

b = 8, b = 2

x^{2} + x - 72 = 0

x, x (100 - x - 2 y) = 0

7 x^{2} - 10 = 0

3 x (x - 2) = 11 x^{2} - 7 x

\frac{a + 3 x ^{2}}{5} = x^{2}

What are the solutions to the equation b^2-10b+10=-6 ?
The solutions to the equation b^2-10b+10=-6 are b=8,b=2
Find the zeros of b^2-10b+10=-6
The zeros of b^2-10b+10=-6 are b=8,b=2