What are the solutions to the equation x^2-4x+53=0 ?

The solutions to the equation x^2-4x+53=0 are x=2+7i,x=2-7i

Find the zeros of x^2-4x+53=0

The zeros of x^2-4x+53=0 are x=2+7i,x=2-7i

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x^{2} - 4 x + 53 = 0

x = 2 + 7 i, x = 2 - 7 i

Solution steps

x^{2} - 4 x + 53 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 53}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 53 : 14 i

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 14 i}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 14 i}{2 \cdot 1}

x = \frac{- ( - 4 ) + 14 i}{2 \cdot 1} : 2 + 7 i

x = \frac{- ( - 4 ) - 14 i}{2 \cdot 1} : 2 - 7 i

The solutions to the quadratic equation are:

x = 2 + 7 i, x = 2 - 7 i

\frac{- 18 a ^{7} b ^{3} + 9 a ^{4} b ^{2} - 63 a ^{2} b}{9 ab}

\frac{1}{3} x = 2 - \frac{2}{3} x

0. 5^{4}

2 x - 1 \leq 3 - x

∣ - 25 x ∣ = 100

What are the solutions to the equation x^2-4x+53=0 ?
The solutions to the equation x^2-4x+53=0 are x=2+7i,x=2-7i
Find the zeros of x^2-4x+53=0
The zeros of x^2-4x+53=0 are x=2+7i,x=2-7i