What are the solutions to the equation 3x^2+51=24x ?

The solutions to the equation 3x^2+51=24x are x=4+i,x=4-i

Find the zeros of 3x^2+51=24x

The zeros of 3x^2+51=24x are x=4+i,x=4-i

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3 x^{2} + 51 = 24 x

x = 4 + i, x = 4 - i

Solution steps

3 x^{2} + 51 = 24 x

Move

24 x

to the left side

3 x^{2} + 51 - 24 x = 0

Write in the standard form

a x^{2} + b x + c = 0

3 x^{2} - 24 x + 51 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 3 \cdot 51}{2 \cdot 3}

Simplify

(- 24)^{2} - 4 \cdot 3 \cdot 51 : 6 i

x_{1, 2} = \frac{- ( - 24 ) \pm 6 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 24 ) + 6 i}{2 \cdot 3}, x_{2} = \frac{- ( - 24 ) - 6 i}{2 \cdot 3}

x = \frac{- ( - 24 ) + 6 i}{2 \cdot 3} : 4 + i

x = \frac{- ( - 24 ) - 6 i}{2 \cdot 3} : 4 - i

The solutions to the quadratic equation are:

x = 4 + i, x = 4 - i

3 x^{2} - 18 x - 15 = 0

x^{2} + 30 x = 231

s^{2} + 4 s + 8

10 n^{2} + 21 n + 9 = 0

1 - x^{2} = - 29

What are the solutions to the equation 3x^2+51=24x ?
The solutions to the equation 3x^2+51=24x are x=4+i,x=4-i
Find the zeros of 3x^2+51=24x
The zeros of 3x^2+51=24x are x=4+i,x=4-i