What are the solutions to the equation p^2-14p+10=-8p ?

The solutions to the equation p^2-14p+10=-8p are p=3+i,p=3-i

Find the zeros of p^2-14p+10=-8p

The zeros of p^2-14p+10=-8p are p=3+i,p=3-i

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p^{2} - 14 p + 10 = - 8 p

p = 3 + i, p = 3 - i

Solution steps

p^{2} - 14 p + 10 = - 8 p

Move

8 p

to the left side

p^{2} - 6 p + 10 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

Simplify

(- 6)^{2} - 4 \cdot 1 \cdot 10 : 2 i

p_{1, 2} = \frac{- ( - 6 ) \pm 2 i}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- ( - 6 ) + 2 i}{2 \cdot 1}, p_{2} = \frac{- ( - 6 ) - 2 i}{2 \cdot 1}

p = \frac{- ( - 6 ) + 2 i}{2 \cdot 1} : 3 + i

p = \frac{- ( - 6 ) - 2 i}{2 \cdot 1} : 3 - i

The solutions to the quadratic equation are:

p = 3 + i, p = 3 - i

5 x^{2} - 2 = 3 - 3 x^{2}

a^{2} + 4 = 13 + 7

- x^{2} - 2 x + 10 = 0

2 x - 7 = - x^{2}

x, f (a + 1) = 4 x^{2} - x + 4

What are the solutions to the equation p^2-14p+10=-8p ?
The solutions to the equation p^2-14p+10=-8p are p=3+i,p=3-i
Find the zeros of p^2-14p+10=-8p
The zeros of p^2-14p+10=-8p are p=3+i,p=3-i