What are the solutions to the equation x^2-4x+148=0 ?

The solutions to the equation x^2-4x+148=0 are x=2+12i,x=2-12i

Find the zeros of x^2-4x+148=0

The zeros of x^2-4x+148=0 are x=2+12i,x=2-12i

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x^{2} - 4 x + 148 = 0

x = 2 + 12 i, x = 2 - 12 i

Solution steps

x^{2} - 4 x + 148 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 148}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 148 : 24 i

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 24 i}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 24 i}{2 \cdot 1}

x = \frac{- ( - 4 ) + 24 i}{2 \cdot 1} : 2 + 12 i

x = \frac{- ( - 4 ) - 24 i}{2 \cdot 1} : 2 - 12 i

The solutions to the quadratic equation are:

x = 2 + 12 i, x = 2 - 12 i

(4 - 3 i) (4 + 3 i)

(x - 2) (x + 5) > 0

16 a^{2} - 40 a + 25

∣ y ∣ = - 9

14 < - 4 a - 8

What are the solutions to the equation x^2-4x+148=0 ?
The solutions to the equation x^2-4x+148=0 are x=2+12i,x=2-12i
Find the zeros of x^2-4x+148=0
The zeros of x^2-4x+148=0 are x=2+12i,x=2-12i