What are the solutions to the equation -10=r^2-10r+12 ?

The solutions to the equation -10=r^2-10r+12 are r=5+sqrt(3),r=5-sqrt(3)

Find the zeros of -10=r^2-10r+12

The zeros of -10=r^2-10r+12 are r=5+sqrt(3),r=5-sqrt(3)

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- 10 = r^{2} - 10 r + 12

r = 5 + 3, r = 5 - 3

Decimal Notation

r = 6.73205 \dots, r = 3.26794 \dots

Solution steps

- 10 = r^{2} - 10 r + 12

Switch sides

r^{2} - 10 r + 12 = - 10

Move

10

to the left side

r^{2} - 10 r + 22 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 22}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 22 = 23

r_{1, 2} = \frac{- ( - 10 ) \pm 2 3}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 10 ) + 2 3}{2 \cdot 1}, r_{2} = \frac{- ( - 10 ) - 2 3}{2 \cdot 1}

r = \frac{- ( - 10 ) + 2 3}{2 \cdot 1} : 5 + 3

r = \frac{- ( - 10 ) - 2 3}{2 \cdot 1} : 5 - 3

The solutions to the quadratic equation are:

r = 5 + 3, r = 5 - 3

x^{2} + 8 x = - 6

- \frac{q ^{2}}{9} + 3 q - 2 = 0

\frac{2 x ^{2} - 3}{4} = 5

x, 4 x^{2} + k x + 9 = 0

15 - x^{2} = 0

What are the solutions to the equation -10=r^2-10r+12 ?
The solutions to the equation -10=r^2-10r+12 are r=5+sqrt(3),r=5-sqrt(3)
Find the zeros of -10=r^2-10r+12
The zeros of -10=r^2-10r+12 are r=5+sqrt(3),r=5-sqrt(3)