20 points! Thank you :)
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Answer:
[tex]\Large \boxed{x=-2 \ \mathrm{and} \ x=3}[/tex]
Step-by-step explanation:
The quadratic expression is given.
[tex]2x^2-2x-12=0[/tex]
Factor the left side of the equation.
[tex]2(x+2)(x-3)=0[/tex]
Set the factors equal to 0.
[tex]x+2=0[/tex]
[tex]x=-2[/tex]
[tex]x-3=0[/tex]
[tex]x=3[/tex]
The two solutions work in the original equation. The solutions make the equation true.
The quadratic expression:
[tex]2 {x}^{2} - 2x - 12[/tex]
★ By split middle term,we get
[tex]2 {x}^{2} - 6x + 4x - 12[/tex]
[tex]2x(x - 3) + 4( x - 3)[/tex]
[tex](2x + 4) (x - 3)[/tex]
[tex]2x + 4 = 0 \: \: and \: \: x - 3 = 0[/tex]
[tex]2 x = - 4 \: \: and \: \: x = 3[/tex]
[tex]x = - 2 \: \: and \: \: x = 3[/tex]
Therefore , the two solution of given quadratic equations is -2 and 3.