Question

solve tha following quadratic equation by comparing the square method
$6 x{}^{2} + x = 2$
​

Answer 1

$\red{\dag}$ Given :-

$\sf{6 {x}^{2} + x = 2}$

$\green{\dag}$ Solution :-

$\sf{6 {x}^{2} + x = 2}$

$:\implies \sf{6x^{2}+x-2=0 }$

$\sf{ : \implies \left(6x^{2}-3x\right)+\left(4x-2\right) } \:$

$\sf{ : \implies 3x\left(2x-1\right)+2\left(2x-1\right) } \:$

$\sf{ : \implies \left(2x-1\right)\left(3x+2\right) } \:$

$\bf \red{ : \implies x=\frac{1}{2} } \\ \bf \red{And, \: \: \bf{ x=-\frac{2}{3} } \: }$

$\fbox{\mathfrak{@ItzPhantom}}$

Answer 2

Step-by-step explanation:

Given:-

6x*2 + x =2

Solution:-

6x*2 + x =2

: 6x*2 + x - 2 =0

: (6x*2 - 3x) + (4x - 2)

: 3x(2x-1) + 2(2x - 1)

: (2x - 1) (3x + 2)

: x = 1/2

And, x = -2/3.

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