What are the solutions to the equation x^2+10x+31=15x+21 ?

The solutions to the equation x^2+10x+31=15x+21 are x= 5/2+i(sqrt(15))/2 ,x= 5/2-i(sqrt(15))/2

Find the zeros of x^2+10x+31=15x+21

The zeros of x^2+10x+31=15x+21 are x= 5/2+i(sqrt(15))/2 ,x= 5/2-i(sqrt(15))/2

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x^{2} + 10 x + 31 = 15 x + 21

x = \frac{5}{2} + i \frac{15}{2}, x = \frac{5}{2} - i \frac{15}{2}

Solution steps

x^{2} + 10 x + 31 = 15 x + 21

Move

21

to the left side

x^{2} + 10 x + 10 = 15 x

Move

15 x

to the left side

x^{2} - 5 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot 10 : 15 i

x_{1, 2} = \frac{- ( - 5 ) \pm 15 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 15 i}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 15 i}{2 \cdot 1}

x = \frac{- ( - 5 ) + 15 i}{2 \cdot 1} : \frac{5}{2} + i \frac{15}{2}

x = \frac{- ( - 5 ) - 15 i}{2 \cdot 1} : \frac{5}{2} - i \frac{15}{2}

The solutions to the quadratic equation are:

x = \frac{5}{2} + i \frac{15}{2}, x = \frac{5}{2} - i \frac{15}{2}

(x - 1) (4 x - 5) = 0

7 x^{2} - 9 = 0

6 x (x - 1) + 4 (x + 4) (x - 1) = (2 x + 2) (x + 4)

x^{2} + 1 = - 99

x^{2} - x - 20 = 0

What are the solutions to the equation x^2+10x+31=15x+21 ?
The solutions to the equation x^2+10x+31=15x+21 are x= 5/2+i(sqrt(15))/2 ,x= 5/2-i(sqrt(15))/2
Find the zeros of x^2+10x+31=15x+21
The zeros of x^2+10x+31=15x+21 are x= 5/2+i(sqrt(15))/2 ,x= 5/2-i(sqrt(15))/2