What are the solutions to the equation 4x^2-52x+169=121 ?

The solutions to the equation 4x^2-52x+169=121 are x=12,x=1

Find the zeros of 4x^2-52x+169=121

The zeros of 4x^2-52x+169=121 are x=12,x=1

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4 x^{2} - 52 x + 169 = 121

x = 12, x = 1

Solution steps

4 x^{2} - 52 x + 169 = 121

Move

121

to the left side

4 x^{2} - 52 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 52 ) \pm ( - 52 ) ^{2} - 4 \cdot 4 \cdot 48}{2 \cdot 4}

(- 52)^{2} - 4 \cdot 4 \cdot 48 = 44

x_{1, 2} = \frac{- ( - 52 ) \pm 44}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 52 ) + 44}{2 \cdot 4}, x_{2} = \frac{- ( - 52 ) - 44}{2 \cdot 4}

x = \frac{- ( - 52 ) + 44}{2 \cdot 4} : 12

x = \frac{- ( - 52 ) - 44}{2 \cdot 4} : 1

The solutions to the quadratic equation are:

x = 12, x = 1

2 (3) + 5

x^{2} y^{2} - 36 y^{4}

3 (x - 2) - 8 x < 44

\frac{2 - 3 i}{3 + i}

∣ 4 (x - 1) + 6 ∣ \leq 10

What are the solutions to the equation 4x^2-52x+169=121 ?
The solutions to the equation 4x^2-52x+169=121 are x=12,x=1
Find the zeros of 4x^2-52x+169=121
The zeros of 4x^2-52x+169=121 are x=12,x=1