How do you work out x^2 + 3x - 10 = 0 (ten points)

can you show the working out please?

Factorise it

3x can split up into 5x and 2x

i.e 3x = 5x - 2x

Place above in equation

x^2 + 3x - 10 = 0

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

x ( x + 5 ) - 2 ( x + 5 ) = 0

(x+5 ) ( x - 2 ) = 0

x + 5 = 0 , x - 2 = 0

x = -5 , x = 2

So x = 2 , -5

use ac method if you can't spot factors straight away

a is number in front of x^2..............in this case +1
c is number constant......................in this case - 10

signs are important

a times c = -10
factors of -10 which will leave +3 in the middle are
+5 times -2
[+5-2 = +3]

now rewrite expression

x^2 -2x + 5x -10 = 0..........this has same value

bracket up in such a way to make a common factor

x[x-2] +5[x -2] = 0

[x-2] is common

Leaves

[x-2][x + 5] = 0

when x-2 = 0
x = +2
when x+5 = 0
x = -5

Solutions

x = +2 or -5

x^2 + 3x - 10
Look at the last number, -10. Use the product and sum method. So basically, what two numbers when multiplied equal -10, but when added equal 3? The answers are 5 and -2. Now, you want to set up something like this:
(x + ?)(x + ?). Put those two numbers in the factor. So, your answer is (x + 5)(x - 2).
Use FOIL to check your answer.

x^2 + 3x - 10 = 0
or x^2 + 5x - 2x - 10 = 0
or x ( x + 5 ) - 2 ( x + 5 ) = 0
or (x+5 ) ( x - 2 ) = 0
either x + 5 = 0 or x - 2 = 0
either x = -5 or x = 2
Hence x = 2 , -5 is the solution to the given problem.

x^2+3x-10 =0


=> x^2 -2x +5x -10 = 0

=> x(x-2) + 5(x-2) = 0

=> (x-2) * (x+5) = 0

or, x = 2 or x = -5

sol. set (2 ;-5)

x^2+3x-10=0

=>x^2+5x-2x-10=0

=>x(x+5)-2(x+5)=0

=>(x+5)(x-2)=0

=>x+5=0 or x-2=0

=>x=-5 or x=2

(x-2)(x+5) = 0

x-2=0
x=2

x+5=0
x=-5

:)

(X+5)(x-2)

x2+5x-2x+10=0

x(x=5)+2(x+5)=0

(x+5)(x+2)=0

x=-5 or -2

its (x+5)(x-2)=0
so x+5=0 ---> x= -5
x-2=0 ---> x=2