how do you solve x^2+5x=3?

Answer 1

x² + 5x - 3 = 0
x = [- 5 ± √(25 + 12)] / 2
x = [- 5 ± √37] / 2
x = 0.54, x = - 5.54 ( to 2 decimal places)

Answer 2

Completing the square

x² + 5x = 3

x² + 5x +_____ = 3 +_____

x² + 5x + 25/4 = 3 + 25/4

(x + 5/2)(x + 5/2) = 12/4 + 25/4

(x + 5/2)² = 37/4

(√x + 5/2)² = ± √37 / √4

x + 5/2 = ± √37 / 2

x + 5/2 - 5/2 = - 5/2 ± √37/2

x = - 5/2 ± √37/2

x = - 5/2 ± 6.08276253 / 2

- - - - - - - - -

Solving for +

x = - 5/2 + 6.08276253 / 2

x = 1.08276253 / 2

x = 0.541381265

- - - - - - -

Solving for -

x = - 5/2 - 6.08276253 / 2

x = - 11.08276253 / 2

x = - 5.541381265

- - - - - - -s-

Answer 3

there is a formula for solving quadratic eq. which is called SHREEDHARACHARYA METHOD
let a general eq. a x^2 + bx + c = 0
then the roots will be, x1 = -b +( b^2 - 4 ac )^1/2 divided by 2a
and x2 = -b - ( b^2 - 4ac )^1/2 divided by 2a
here a= 1, b= 5, c= -3
x1 = -5 + ( 5^2- 4*1*-3 )^1/2 divided by 2*1
= -5 + ( 37)^1/2 / 2
= 0.5413
x2 = -5 - (37)^1/2 / 2
= - 5.54138

Answer 4

x^2+5x-3 = 0
delta = b^2-4ac = 5^2+4*3 = 25+12 = 37
x1=[-5-sqrt(37)]/2
x2=[-5+sqrt(37)]/2

Answer 5

you break the middle part having coefficient of x so that equation can be factorized(common parts come togeather). Here it cannot be done. So use discriminant method to find the D and x.

x= [-b+-sqrt(b^2-4ac)]/2a
so, x=[-5+-sqrt(25+12)]/2

solve it. this is the answer