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Select the two values of x that are roots of this equation x^2+3x-5=0

2 Answers

2 votes

x = -(3-sqa(29))/2

x= -(3+sqa(29))/2

decimal form

x = 1.193, -4.1923

User Philly
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6.1k points
3 votes

Answer : The two values of 'x' are, 1.192 and -4.192

Step-by-step explanation :

The given expression is,


x^2+3x-5=0

To solve this problem we are using quadratic formula.

The general quadratic equation is,


ax^2+bx+c=0

Formula used :


x=(-b\pm √(b^2-4ac))/(2a)

Now we a have to solve the above equation and we get the value of 'x'.


x^2+3x-5=0

a = 1, b = 3, c = -5


x=(-b\pm √(b^2-4ac))/(2a)


x=(-(3)\pm √((3)^2-4* 1* (-5)))/(2* 1)


x=(-3+√((3)^2-4* 1* (-5)))/(2* 1)


x=1.192

and,


x=(-3-√((3)^2-4* 1* (-5)))/(2* 1)


x=-4.192

Therefore, the values of 'x' are 1.192 and -4.192

User Erdogan
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6.1k points