If it's not what You are looking for type in the equation solver your own equation and let us solve it.
15x^2+150x-18=0
a = 15; b = 150; c = -18;
Δ = b2-4ac
Δ = 1502-4·15·(-18)
Δ = 23580
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{23580}=\sqrt{36*655}=\sqrt{36}*\sqrt{655}=6\sqrt{655}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-6\sqrt{655}}{2*15}=\frac{-150-6\sqrt{655}}{30} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+6\sqrt{655}}{2*15}=\frac{-150+6\sqrt{655}}{30} $
| -20=3(w-8)-5w | | 9{10x+4}-2=52 | | 14x=7-35 | | 32.1y+3.1+2.4-8.2=34.5y-5.1 | | x+x/3+x+5=19 | | 7(1+2v)=105 | | 36=7v+3(v+2) | | 3n+6+5=11 | | 2y+12y-3=8y+27 | | -2v+3=-7v+16 | | 17x-45=6 | | 5m-24=48 | | -18-x=2 | | 2(3u+4)=26 | | 1/3(9x+2)=2x+1/2 | | 7v+8-10v=-19 | | 3(y+4)−4(3y+2)=−2−10y | | 22q+14=-5q-8 | | 54=10u-4u | | 5x+8/2=3x-1 | | -5.4=-13.3+x | | 1/3(9x+7)=2x+1/2 | | 3u+3u=42 | | 4y-6+2(3y+6)=-6(y+2) | | -47+9x=2x+65 | | -25+4x=37+6x | | 2−5x=3(2−3x) | | p(3)=2.07 | | -2x-121=199+14x | | H(x)=3x^2+5x+2 | | 180=100+(x+12)+x | | -94+x=126+11x |