What are the solutions to the equation x^2-14x+48=-10 ?

The solutions to the equation x^2-14x+48=-10 are x=7+3i,x=7-3i

Find the zeros of x^2-14x+48=-10

The zeros of x^2-14x+48=-10 are x=7+3i,x=7-3i

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x^{2} - 14 x + 48 = - 10

x = 7 + 3 i, x = 7 - 3 i

Solution steps

x^{2} - 14 x + 48 = - 10

Move

10

to the left side

x^{2} - 14 x + 58 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 58}{2 \cdot 1}

Simplify

(- 14)^{2} - 4 \cdot 1 \cdot 58 : 6 i

x_{1, 2} = \frac{- ( - 14 ) \pm 6 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 6 i}{2 \cdot 1}, x_{2} = \frac{- ( - 14 ) - 6 i}{2 \cdot 1}

x = \frac{- ( - 14 ) + 6 i}{2 \cdot 1} : 7 + 3 i

x = \frac{- ( - 14 ) - 6 i}{2 \cdot 1} : 7 - 3 i

The solutions to the quadratic equation are:

x = 7 + 3 i, x = 7 - 3 i

2 x^{2} + 14 x - 120 = 0

6 x^{2} = 3 x - 9

x, \frac{x ^{2}}{2} + \frac{y ^{2}}{3} + \frac{z ^{2}}{4} = 1

x^{2} = e^{4}

s^{2} - 2 s - 2 = 0

What are the solutions to the equation x^2-14x+48=-10 ?
The solutions to the equation x^2-14x+48=-10 are x=7+3i,x=7-3i
Find the zeros of x^2-14x+48=-10
The zeros of x^2-14x+48=-10 are x=7+3i,x=7-3i