How do you solve using the quadratic form? 2x^2+3x-3=-3x-4

3 Answers
May 8, 2018

Do you mean the quadratic formula?
x=-0.1771243445 or x=-2.822875656x=0.1771243445orx=2.822875656

Explanation:

2x^2+3x-3=-3x-42x2+3x3=3x4 collect like terms on the LHS

2x^2+6x+1=02x2+6x+1=0

a=2, b=6 and c=1a=2,b=6andc=1
x=[-6\pmsqrt(6^2-[4xx2xx1])]/[2xx2]x=6±62[4×2×1]2×2

x=[-6\pmsqrt(36-8)]/[4]x=6±3684

x=[-6\pmsqrt28]/[4]x=6±284

x=[-6\pm2sqrt7]/[4]x=6±274

x=(-3+sqrt7)/2x=3+72 or x=(-3-sqrt7)/2x=372

x=-0.1771243445 or x=-2.822875656x=0.1771243445orx=2.822875656

May 8, 2018

2 x^2+3 x-3=-3 x-42x2+3x3=3x4

Explanation:

a x^2+b x+c=0ax2+bx+c=0
x_(1,2)=(-b +- \sqrt(b^2-4*a*c))/(2*a)x1,2=b±b24ac2a

Write your equation in the form: a x^2+b x+c=0ax2+bx+c=0
2 x^2+3 x-3=-3 x-42x2+3x3=3x4 | +3x+3x
2 x^2+3 x-3 +3x=-3 x-4 +3x2x2+3x3+3x=3x4+3x |+4+4
2 x^2+6 x-3+4=-4 +42x2+6x3+4=4+4
2 x^2+6 x+1=02x2+6x+1=0

a=2a=2
b=6b=6
c=1c=1

x_(1,2)=(-6 +- \sqrt(6^2-4*2*1))/(2*2)=x1,2=6±6242122=
=(-6 +- \sqrt(36-8))/(4)==6±3684=
=(-6 +- \sqrt(28))/(4)==6±284=
=(-6 +- \sqrt(2^2*7))/(4)==6±2274=
=(-6 +- 2* \sqrt(7))/(4)==6±274=
=(-3 +- \sqrt(7))/(2)=3±72

x_1=(-3 + \sqrt(7))/(2)x1=3+72
x_2=(-3 - \sqrt(7))/(2)x2=372

May 8, 2018

x=(-3+sqrt7)/2 or x=(-3-sqrt7)/2x=3+72orx=372

Explanation:

Here,

2x^2+3x-3=-3x-42x2+3x3=3x4

=>2x^2+3x-3+3x+4=02x2+3x3+3x+4=0

=>2x^2+6x+1=02x2+6x+1=0

Comparing with ax^2+bx+c=0ax2+bx+c=0,

a=2,b=6 and c=1a=2,b=6andc=1

So,

triangle=b^2-4ac=36-4(2)(1)=36-8=28=b24ac=364(2)(1)=368=28

sqrt(triangle)=2sqrt7=27

:.x=(-b+-sqrt(triangle))/(2a)=(-6+-2sqrt7)/(2(2))=(-3+-sqrt7)/2

Hence,

x=(-3+sqrt7)/2 or x=(-3-sqrt7)/2