What are the solutions to the equation r^2-5r+4=0 ?

The solutions to the equation r^2-5r+4=0 are r=4,r=1

Find the zeros of r^2-5r+4=0

The zeros of r^2-5r+4=0 are r=4,r=1

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r^{2} - 5 r + 4 = 0

r = 4, r = 1

Solution steps

r^{2} - 5 r + 4 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 4 = 3

r_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 1}, r_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 1}

r = \frac{- ( - 5 ) + 3}{2 \cdot 1} : 4

r = \frac{- ( - 5 ) - 3}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

r = 4, r = 1

7 x^{2} - 3 x - 972 = 0

3 x^{2} - 30 x - 11 = 6 x

\frac{1}{7} \cdot 5

- 3 x + 20 > 2 x + 5

∣ 3 a + 9 ∣ = ∣ 4 a - 1 ∣

What are the solutions to the equation r^2-5r+4=0 ?
The solutions to the equation r^2-5r+4=0 are r=4,r=1
Find the zeros of r^2-5r+4=0
The zeros of r^2-5r+4=0 are r=4,r=1