What are the solutions to the equation (5-r)(1-r)+20=0 ?

The solutions to the equation (5-r)(1-r)+20=0 are r=3+4i,r=3-4i

Find the zeros of (5-r)(1-r)+20=0

The zeros of (5-r)(1-r)+20=0 are r=3+4i,r=3-4i

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(5 - r) (1 - r) + 20 = 0

r = 3 + 4 i, r = 3 - 4 i

Solution steps

(5 - r) (1 - r) + 20 = 0

Expand

(5 - r) (1 - r) + 20 : r^{2} - 6 r + 25

r^{2} - 6 r + 25 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 25}{2 \cdot 1}

Simplify

(- 6)^{2} - 4 \cdot 1 \cdot 25 : 8 i

r_{1, 2} = \frac{- ( - 6 ) \pm 8 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 6 ) + 8 i}{2 \cdot 1}, r_{2} = \frac{- ( - 6 ) - 8 i}{2 \cdot 1}

r = \frac{- ( - 6 ) + 8 i}{2 \cdot 1} : 3 + 4 i

r = \frac{- ( - 6 ) - 8 i}{2 \cdot 1} : 3 - 4 i

The solutions to the quadratic equation are:

r = 3 + 4 i, r = 3 - 4 i

4 x^{2} - x - 1 = 5

36 x^{2} - 49 = 0

2 x^{2} - 5 x = 6

40 = 30 t - 5 t^{2}

f, f (1 - x) = x^{2} - 2 x

What are the solutions to the equation (5-r)(1-r)+20=0 ?
The solutions to the equation (5-r)(1-r)+20=0 are r=3+4i,r=3-4i
Find the zeros of (5-r)(1-r)+20=0
The zeros of (5-r)(1-r)+20=0 are r=3+4i,r=3-4i