If it's not what You are looking for type in the equation solver your own equation and let us solve it.
4x^2-3=12x
We move all terms to the left:
4x^2-3-(12x)=0
a = 4; b = -12; c = -3;
Δ = b2-4ac
Δ = -122-4·4·(-3)
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{3}}{2*4}=\frac{12-8\sqrt{3}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{3}}{2*4}=\frac{12+8\sqrt{3}}{8} $
| –9q=–8q−5 | | 8x-1=5x+47 | | –2n+8=7n−10 | | 6k+2=36 | | 6=1x= | | 5/12=b/5= | | (10-x)(x-3)=0 | | 2x-9=4(x-3)+19 | | -3x-10=32x= | | (x+4)/10=(x-2)/5 | | 5x+9=65x= | | 12x-4=4x-6+82 | | 7x^2-x=3 | | W=4w-96 | | 4x-6x+8x=40 | | 4(m+4)=m+ | | m^2-4m+6=0 | | (3,–8)m=–5 | | 10p+33=p+60 | | x^2-14x-284=-2x^2 | | 9+22x=10x+7+74 | | 9+22x=10x+7+84 | | (D^4-4D^3+7D^2+4D+6)y=0 | | P-32+2p-46=2(p-5) | | 3.5+0.65a=16-0.3a | | 1,995=(x+19)25 | | 12x-5=7x+45 | | 23=y/4-10 | | 1870=30a+17 | | (X+5)^2-2(x-5)-24=0 | | (X+5)-2(x-5)^2-24=0 | | x^2−4x+4=20 |