guys i can't understand please tell me how to solve it????? in the form of quadratic formula
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Answered by
1
Hey mate!!!!!
Here is your answer.
2(x-1)^2=2x+3
2x^2+2x-1=0
D=b^2-4ac
=2^2-4(2)(-1)
=4+8
=12
It has two real and distinct roots.
x=-b+-√b^2-4ac/2a
=-2+-√12/2(2)
=-2+-2√3/4
x=-2+2√3/4 and x=-2-2√3/4.
Hope it helps!!!!!
Here is your answer.
2(x-1)^2=2x+3
2x^2+2x-1=0
D=b^2-4ac
=2^2-4(2)(-1)
=4+8
=12
It has two real and distinct roots.
x=-b+-√b^2-4ac/2a
=-2+-√12/2(2)
=-2+-2√3/4
x=-2+2√3/4 and x=-2-2√3/4.
Hope it helps!!!!!
Palak2408:
Hii! Thank you for selecting my answer as the brainliest.
welcome
simu plzzz massage me in inbox mujhe thum se baat karna hai please
i hope you understood
please
simu
Answered by
1
2(x square - 1)= 2x +3
2x square - 2 = 2x +3
2xsquare-2x=3+2
x = 5
2x square - 2 = 2x +3
2xsquare-2x=3+2
x = 5
thanks for your help
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