What are the solutions to the equation x^2-(5-3x)=x+10 ?

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

Find the zeros of x^2-(5-3x)=x+10

The zeros of x^2-(5-3x)=x+10 are x=3,x=-5

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x^{2} - (5 - 3 x) = x + 10

x = 3, x = - 5

Solution steps

x^{2} - (5 - 3 x) = x + 10

Expand

x^{2} - (5 - 3 x) : x^{2} - 5 + 3 x

x^{2} - 5 + 3 x = x + 10

Move

10

to the left side

x^{2} + 3 x - 15 = x

Move

x

to the left side

x^{2} + 2 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 15) = 8

x_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 8}{2 \cdot 1}, x_{2} = \frac{- 2 - 8}{2 \cdot 1}

x = \frac{- 2 + 8}{2 \cdot 1} : 3

x = \frac{- 2 - 8}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 3, x = - 5

x^{2} - 9 x = x + 24

2 x^{2} + 5 x = 0

8 x^{2} - 16 x - 4 = 0

(2 x - 6)^{2} - 64 = 0

y, 2 x - y^{2} = 7

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