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

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

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

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

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

x = a + 2, x = - 5

Solution steps

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

Expand

a (x + 5) : a x + 5 a

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

Switch sides

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

Subtract

a x + 5 a

from both sides

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

Simplify

x^{2} + (3 - a) x - 10 - 5 a = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 3 - a ) \pm ( 3 - a ) ^{2} - 4 \cdot 1 \cdot ( - 10 - 5 a )}{2 \cdot 1}

Simplify

(3 - a)^{2} - 4 \cdot 1 \cdot (- 10 - 5 a) : a + 7

x_{1, 2} = \frac{- ( 3 - a ) \pm ( a + 7 )}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( 3 - a ) + a + 7}{2 \cdot 1}, x_{2} = \frac{- ( 3 - a ) - ( a + 7 )}{2 \cdot 1}

x = \frac{- ( 3 - a ) + a + 7}{2 \cdot 1} : a + 2

x = \frac{- ( 3 - a ) - ( a + 7 )}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = a + 2, x = - 5

2 x^{2} - 3 = 29

5 = (b + 1)^{2} + b^{2}

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

(x - 6) (x - 6) = 0

36 = m^{2}

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